Kassuno wrote: Find all the pairs of integers $(k,n)$ such that, $$k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})$$ Shouldn't there not be a comma after "such that"? There also needs to be a period at the end. I think the question should look like this. Kassuno wrote: Find all the pairs of integers $(k,n)$ such that$$k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1}).$$

TwinPrime wrote: Kassuno wrote: Find all the pairs of integers $(k,n)$ such that, $$k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})$$ Shouldn't there not be a comma after "such that"? There also needs to be a period at the end. I think the question should look like this. Kassuno wrote: Find all the pairs of integers $(k,n)$ such that$$k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1}).$$ That doesn't really matter, I guess?

ImbecileMathImbaTation wrote: That doesn't really matter, I guess? For things like the contest collection, which should match the actual formatting, I believe so.

TwinPrime wrote: ImbecileMathImbaTation wrote: That doesn't really matter, I guess? For things like the contest collection, which should match the actual formatting, I believe so. Oh i see

TwinPrime wrote: ImbecileMathImbaTation wrote: That doesn't really matter, I guess? For things like the contest collection, which should match the actual formatting, I believe so. I think that the expression $(2^n-1)(2^n-2)...(2^n-2^{n-1})$ is more important than a missed comma. This can be confused to $(2^n-1)(2^n-2)(2^n-3)...(2^n-2^{n-1}).$

We first compare the $v_2$'s of each side. The left-hand side has $v_2(\text{LHS})=\sum_{i=1}^{\infty} \lfloor\frac{k}{2^i} \rfloor \le k$, whereas the right hand side has exactly $1+2+\cdots +(n-1)=\frac{n(n-1)}{2}$. Hence $\boxed{k \ge \frac{n(n-1)}{2}}$. Now we compare the $v_3$'s of each side. Note that $v_3(2^{2j}-1)=1+v_3(j)$ by the LTE lemma, so $v_3(k!)=\lfloor\frac{n}{2}\rfloor+v_3(n!)$. Since $v_3(k!)-v_3(n!)\ge \frac{k-n}{3}-2$, so we get $\boxed{k\le n+\frac{3n}{2}+6}$. Comparing the two inequalities give finite cases(i.e. something like $n\le 7$), so we're done.

We will use abbrevations $RS$ and $LS$ for left and right side respectively. It is a common fact that $\nu_2(k!)\leqslant k-1$. On the other hand, $\nu_2(RS)=0+1+2+\ldots+n-1=\frac{n^2-n}{2}$. This implies $\frac{n^2-n+2}{2}\leqslant k$, and, taking factorials, we get $\frac{n^2-n+2}{2}! \leqslant (2^n-1)\ldots(2^n-2^{n-1}) < (2^n)^n=2^{n^2}$. Using a well-known bound $x!\geqslant (\frac{x}{3})^x$, we obtain $$(\frac{n^2-n+2}{6})^{\frac{n^2-n+2}{2}} < 2^{n^2}.$$ Taking base two logarithm, we get $$\frac{n^2-n+2}{2} \log_2 \frac{n^2-n+2}{6} < n^2.$$ For $n>7$, the expression under the logarithm is bigger than $8$, and consequently the left side is bigger that the right side. Therefore, $n\leqslant7$. It can now be easily checked that $(n,k) \in \{(1,0),(1,1),(2,3)\}$ are the only solutions.

$\text{RHS} = (2^n - 1)\cdot 2^{n-1} (2^{n-1} - 1)\cdot (2^{n-1} - 2) \cdot \ldots \cdot (2^{n-1} - 2^{n-2}).$ When $2^{n-1} - 2^{n-2} = 2^{n-2} > 1,$ i.e. $n > 2,$

I was sure that there will be a Number Theory problem on Day 2. By the way, How much time did the contestants generally take to solve the P4 of the Day 2 ?

Nice problem!... much better than day 1- First, let's write $(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})$ as $\frac{(2^n -1)!}{(2^{n-1}-1)!}$. Now, use $v_2(n!)=\frac{n-S_2(n)}{2-1}$ to get $v_2((2^n -1)!)=2^n - n -1$ and $v_2((2^{n-1})!)=2^{n-1}-n$, thus $v_2((2^n-1)\cdots(2^n-2^{n-1})=2^{n-1}-1$. And now, we realize that $v_2((2^{n-1})!)=2^{n-1}-1$, and the problem becomes very easy from here (basically $2^{n-1} \leq k \leq 2^{n-1}+1$, and as we know $k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1}) => (2^n-1)(2^n-2)\cdots(2^n-2^{n-1})=(2^{n-1})(2^{n-1}-1)) \cdots 1$ which is trivially false for $n>1$ as there are $2^{n-1}$ terms on both sides, and the other case is very similar, though a little more complicated-the answers will be $n=1,2$ for $k=1,3$ respectively). QED. Honestly, I feel this is a much better problem and way more fun to solve than P1, although a little on the easier side if you know Legendre's formula(I'm not sure if this is what it's called though)...

Kassuno wrote: Find all the pairs of integers $(k,n)$ such that $$k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1}).$$ The actual problem says positive integers. The solution is just bounding. By Legendre, $\nu_2(k!) < n$. We clearly have \[ \nu_2 (k!) = \nu_2 \left( \prod_{i=0}^{n-1} (2^n - 2^i) \right) = \frac{(n(n-1)}{2} . \]Thus $k! > (\frac{n(n-1)}{2})!$. On the other hand, though, \[ \prod_{i=0}^{n-1} (2^n - 2^i) > 2^{n^2}. \]It is easy to see that these two facts are impossible for sufficiently large $n$. In fact $n=6$ will suffice since \begin{align*} 15! &= 2 \cdot 3 \cdots 15 \\ &> 2^2 \cdot 4^4 \cdot 8^4 \cdot 12 \cdot 13 \cdot 14 \cdot 15 \\ &= 2^{22} \cdot 132 \cdot 210 \\ &> 2^{22} \cdot 2^7 \cdot 2^7 \\ &= 2^{36} = 2^{6^2} \end{align*}Then we induct up: \[ \left(\frac{n(n+1)}{2}\right)! > 2^{n^2} \cdot 6^{n} > 2^{(n+1)^2} \]for all $n \geq 6$. Thus we only need to consider when $n = 6$. By sheer computation we discover that $(k,n)=(1,1),(3,2)$ are solutions, but when $n=3,4$, there are none. To show that $n=5$ has no solution, note that the right handside is divisible by 31, so then $k! > 31! >2^{30} > 2^{25} $, so we are done.

Here's what I found during the test: Legendre gives $$v_2(k!)<k$$But we also have: $$v_2(RHS)=0+1+...+n-1=\dfrac{n(n-1)}{2}$$Then we must have $k>\dfrac{n(n-1)}{2}$ Therefore $$2^{n^2}>RHS=k!>\bigg(\dfrac{n(n-1)}{2}\bigg)! \ge 1.2.2.4.4.4.4.8.8.8.8.8...8$$$$=2^{2+8+3((n(n-1)/2-7) }$$So we must have : $$n^2\ge 2+8+3(n(n-1)/2-7)$$This holds only for n<7. We can check $n\le 6$ by hand.

Note that n(n-1)/2=v2(k!)<k but (n(n-1)/2)!>2^(n^2) for n>6 by induction hence (k,n)=(2,1);(2,3)

ubermensch wrote: Nice problem!... much better than day 1- Now, use $v_2(n!)=\frac{n-S_2(n)}{2-1}$ to get $v_2((2^n -1)!)=2^n - n -1$ and $v_2((2^{n-1})!)=2^{n-1}-n$, thus $v_2((2^n-1)\cdots(2^n-2^{n-1})=2^{n-1}-1$. . What is $S_2(n)$, bro?

Well i think i found another solution... First rewrite $k!=(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})=\frac{(2^n -1)!}{(2^{n-1}-1)!}$. Also, $v_2(\frac{(2^n -1)!}{(2^{n-1}-1)!})=v_2(2^n-1)!-v_2(2^{n-1}-1)!$. But this equals $2^n-1-n-(2^{n-1}-1-n+1)=2^n-2^{n-1}-1$. Note that we used $S_2(2^n-1)=n$ .But actually, computing in a different way too yields $v_2(k!)=\frac{n^2-n}{2}$. So we have $2^{n}-2^{n-1}-1=2^{n-1}-1 = \frac{n^2-n}{2} \implies 2^n-2=n^2-n$, which implies $2^n-n^2+n=2$. But this happens only when $n=1,2,3$. When $n=1, k!=1 \implies (n,k)=(1,1)$ is a solution. When $n=2, k!=3 \cdot 2 \implies (n,k)=(2,3)$ is another solution. But when $n=3, k!=7 \cdot 6 \cdot 5 \cdot 4$ and this is a contradiction. Hence only solutions are $\boxed{(n,k)=(1,1),(2,3)}$

$v_2(\text{LHS})=\sum_{i=1}^{\infty} \lfloor\frac{k}{2^i} \rfloor \le \sum_{i=1}^{\infty}\frac{k}{2^i} \le k$ and $v_2(\text{RHS}) = \frac{n(n-1)}{2}$. Now we prove with induction that $$(\frac{n^2-n+2}{2})! >2^{\frac{n(n-1)}{2}}(2^n-1)\cdots(2^1-1) $$for $n \ge 6$ Then the equality can not hold. Assume that it is true for $n$ and we prove for $n+1$ Hence it is sufficient to prove that $$(n^2-n+4)(n^2-n+6)\cdots(n^2+n+2) \ge 2^{2n}(2^{n+1}-1)$$Take the obvious inequalities $(n^2-n+4) \ge 2n$ , $(n^2-n+6) \ge 2(n+1)$ , ... and $n^2+n+2 \ge 4n-2$ So it is enough to prove that $n(n+1)\cdots(2n-1) \ge 2^{n+1} -1 $ which is true (even for $n\ge 3$) That means that the equality can not hold for $n \ge 6$ and we can check that $(n,k) \in \{(1,1),(2,3)\}$ are the only solutions

The answer is $(n,k) =(1,1)$ and $(n,k) = (2,3)$ which work. Let $A = \prod_i (2^n-2^k)$, and assume $A = k!$ for some $k \ge 3$. Recall by exponent lifting that \[ \nu_3(2^t-1) = \begin{cases} 0 & t \text{ odd} \\ 1 + \nu_3(t) & t \text{ even}. \end{cases} \]Consequently, we can compute \begin{align*} k > \nu_2(k!) &= \nu_2(A) = 1 + 2 + \dots + (n-1) = \frac{n(n-1)}{2} \\ \frac k3 \le \nu_3(k!) &= \nu_3(A) = \left\lfloor \frac n2 \right\rfloor + \left\lfloor \frac n6 \right\rfloor + \dots < \frac 34n. \end{align*}where the very first inequality can be justified say by Legendre's formula $\nu_2(k!) = k - s_2(k)$. In this way, we get \[ \frac 94 n > k > \frac{n(n-1)}{2} \]which means $n \le \frac{11}{2}$; a manual check then shows the solutions we claimed earlier are the only ones.

I think this problem is $k!=(2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}).$ Solution: $$v_2((2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}))=1+2+\cdots+(n-1)=\dfrac{n(n-1)}{2},$$By Legendre formula, $v_2(k!)=\sum\limits_{j=1}^{+\infty}[\dfrac{k}{2^j}]<k(\dfrac{1}{2}+\dfrac{1}{4}+\cdots)=k$. It is clear that $3 \mid 2^x-1 \Leftrightarrow 2|x$, by LTE Lemma: $v_3(4^x-1)=1+v_3(3)$. We notice: $$v_3((2^n-1)(2^{n-1}-1)\cdots(2-1)) =[\frac{n}{2}]+v_3([\frac{n}{2}]!) <[\frac{n}{2}]+[\frac{n}{2}](\dfrac{1}{3}+\dfrac{1}{9}+\cdots) =\dfrac{3}{2}[\frac{n}{2}].$$$$v_3(k!) \geq [\dfrac{k}{3}]>\dfrac{k}{3}-1.$$Then, $\dfrac{k}{3}-1<\dfrac{3}{2}[\frac{n}{2}] \leq \dfrac{3}{4}n$, link with $\dfrac{n(n-1)}{2}<k$: $\dfrac{n(n-1)}{2}<k<\dfrac{9}{4}n+3$, $2n^2-11n-12<0$, $1 \leq n<6$, then we can know $(k,n)=(1,1), (3,2)$.

ubermensch wrote: First, let's write $(2^n-1)(2^n-2)\cdots(2^n-2^{n-1})$ as $\frac{(2^n -1)!}{(2^{n-1}-1)!}$. That's wrong. WypHxr wrote: I think this problem is $k!=(2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}).$ Solution: $$v_2((2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}))=1+2+\cdots+(n-1)=\dfrac{n(n-1)}{2},$$By Legendre formula, $v_2(k!)=\sum\limits_{j=1}^{+\infty}[\dfrac{k}{2^j}]<k(\dfrac{1}{2}+\dfrac{1}{4}+\cdots)=k$. It is clear that $3 \mid 2^x-1 \Leftrightarrow 2|x$, by LTE Lemma: $v_3(4^x-1)=1+v_3(3)$. We notice: \begin{displaymath} \begin{aligned} v_3((2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}))&=v_3((2^n-1)(2^{n-1}-1)\cdots(2-1)) &=[\frac{n}{2}]+v_3([\frac{n}{2}]!) &<[\frac{n}{2}]+[\frac{n}{2}](\dfrac{1}{3}+\dfrac{1}{9}+\cdots) &=\dfrac{3}{2}[\frac{n}{2}]. \end{aligned} \end{displaymath} $$v_3(k!) \geq [\dfrac{k}{3}]>\dfrac{k}{3}-1.$$Then, $\dfrac{k}{3}-1<\dfrac{3}{2}[\frac{n}{2}] \geq \dfrac{3}{4}n$, link with $\dfrac{n(n-1)}{2}<k$: $\dfrac{n(n-1)}{2}<k<\dfrac{9}{4}n+3$, $2n^2-11n-12<0$, $1 \leq n<6$, then we can know $(k,n)=(1,1), (3,2)$. That's the correct statement.

OG solution, probably the shortest: $v_2{k!}<k$ $v_2{RHS}$ = n(n-1)/2 thus $ n(n-1)/2 <k$ , now RHS<$2^{n(n-1)}$ thus $ k!$<$2^{n(n-1)}$ ($H_k$ denotes k th harmonic number) now $2^{n(n-1)}$> $k!$> ${\frac{k}{H_k}}^{k}$>${\frac{k}{ln(k)+1}}^{k}$ > $4^{k}$ if ($k >=16$ ) thus $n(n-1)/2>k$ which is a contradiction thus, $k<16$ now it is just hit and trial to get 1,1 and 2,3 (n,k) as only solution

please verify someone

sansgankrsngupta wrote: OG solution probably the shortest: $v_2{k!}<k$ $v_2{RHS}$ = n(n-1)/2 thus $ n(n-1)/2 <k$ , now RHS<$2^{n(n-1)}$ thus k!<$2^{n(n-1)}$ ($H_k$ denotes k th harmonic number now $2^{n(n-1)}$> k!> ${\frac{k}{H_k}}^{k}$>${\frac{k}{ln(k)+1}}^{k}$ > $4^{k}$ if ($k >=16$ ) thus n(n-1)/2>k which is a contradiction thus, k<16 now it is just hit and trial to get 1,1 and 2,3 (n,k) as only solutions Can you Latex it properly ,Please. You can use AI tool to do it. It's really hard to read. @Below ,plz edit the latex.

Safal wrote: sansgankrsngupta wrote: OG solution probably the shortest: $v_2{k!}<k$ $v_2{RHS}$ = n(n-1)/2 thus $ n(n-1)/2 <k$ , now RHS<$2^{n(n-1)}$ thus k!<$2^{n(n-1)}$ ($H_k$ denotes k th harmonic number now $2^{n(n-1)}$> k!> ${\frac{k}{H_k}}^{k}$>${\frac{k}{ln(k)+1}}^{k}$ > $4^{k}$ if ($k >=16$ ) thus n(n-1)/2>k which is a contradiction thus, k<16 now it is just hit and trial to get 1,1 and 2,3 (n,k) as only solutions Can you Latex it properly ,Please. You can use AI tool to do it. It's really hard to read. I Edited probably now easy to read

\documentclass{article} \usepackage{amsmath} \begin{document} \[ v_2(k!) < k \] \[ v_2(\text{RHS}) = \frac{n(n-1)}{2} \] thus \[ \frac{n(n-1)}{2} < k \] now \(\text{RHS} < 2^{n(n-1)}\) thus \[ k! < 2^{n(n-1)} \] (H$_k$ denotes the $k$-th harmonic number) now \[ 2^{n(n-1)} > k! > \left(\frac{k}{H_k}\right)^{k} > \left(\frac{k}{\ln(k)+1}\right)^{k} > 4^k \] if \( k \geq 16 \) thus \[ \frac{n(n-1)}{2} > k \]which is a contradiction thus, \( k < 16 \) now it is just hit and trial to get (1, 1) and (2, 3) as the only solutions for \( (n, k) \) \end{document}

OG solution

Safal wrote: sansgankrsngupta wrote: OG solution probably the shortest: $v_2{k!}<k$ $v_2{RHS}$ = n(n-1)/2 thus $ n(n-1)/2 <k$ , now RHS<$2^{n(n-1)}$ thus k!<$2^{n(n-1)}$ ($H_k$ denotes k th harmonic number now $2^{n(n-1)}$> k!> ${\frac{k}{H_k}}^{k}$>${\frac{k}{ln(k)+1}}^{k}$ > $4^{k}$ if ($k >=16$ ) thus n(n-1)/2>k which is a contradiction thus, k<16 now it is just hit and trial to get 1,1 and 2,3 (n,k) as only solutions Can you Latex it properly ,Please. You can use AI tool to do it. It's really hard to read. @Below ,plz edit the latex. I have edited the latex, I hope now you can read

We check $v_2$ and $v_3$ :- \begin{align*} v_2((2^n-1)(2^n-2)(2^n-2^2)\cdots(2^n-2^{n-1}))&=v_2((2^\frac{n(n-1)}{2})\cdot (2^{n-1}-1)\cdot (2^{n-2}-1) \cdots (2-1))\\&=\frac{n(n-1)}{2}\\v_2({k!})&=\Biggl\lfloor \frac{k}{2} \Biggr\rfloor + \Biggl\lfloor \frac{k}{4} \Biggr\rfloor + \cdots \le k \\& \implies k\ge \frac{n(n-1)}{2} \end{align*}Now, from LTE, we have that $\nu_3(2^{2m} - 1) = \nu_3(3m)$ if n is odd then we get $\nu_3(2^{n} - 1) =0$. Also note that $v_3({2^m-2^n})=v_3({2^{m-n}-1})$ \begin{align*} v_3(k!) &= v_3((2^n-1)(2^{n-1} - 1) \cdots 1) \\ &= v_3(2^n -1)v_3(2^{n-1} - 1) \cdots v_3(1) \\ &= \lfloor \tfrac{n}{2} \rfloor + v_3(\lfloor \tfrac{n}{2} \rfloor !)\le \frac{n}{2}+\frac{n}{6}+\ldots=\frac{3}{4}n \geq \Biggl\lfloor\frac{k}{3}\Biggl\rfloor> \frac{k-3}{3} \\& \implies \frac{9}{4}n+3>k\ge \frac{n(n-1)}{2} \implies n\leq 6 \end{align*}By checking manually we see that $(k,n)=\boxed{(1,1),(3,2)}$

We claim that the only answers are $(k,n)=(1,1)$ and $(3,2)$, which clearly work. We can test $n=3$ and $n=4$ to see they do not have an integer solution for $k$. Now note that the right side rewrites as $2^{\frac{n(n-1)}{2}}\cdot\prod_{i=1}^{n}(2^i-1)$. Since $2^5-1=31$, every $i$ divisible by $5$ contributes at least one factor of $31$. Thus, $\nu_{31}\left(\prod_{i=0}^{n-1} (2^n-2^i)\right)\geq \left\lfloor \frac n5 \right \rfloor$. Since the order of $2\pmod{11}$ is $10$ and $\nu_{11}(1023)=1$, using LTE we have that $\nu_{11} \left(\prod_{i=0}^{n-1} (2^n-2^i)\right) = \left\lfloor \frac {n}{10} \right \rfloor+\left\lfloor \frac {n}{110} \right \rfloor+\left\lfloor \frac {n}{1210} \right \rfloor+\dots$, which is always less than $\left\lfloor \frac {n}{5} \right \rfloor$. However, $\nu_{11}(k!)\geq \nu_{31}(k!)$ for all $k$, thus the two sides cannot be equal for $n>4$, and there are no more solutions.

We claim that the only pairs of solutions are $(k,n)=(1,1)$ and $(3,2)$. It is not hard to see that these solutions work, now we shall show that they are the only ones. First note that comparing $\nu_2$ of both sides we have, \begin{align*} \frac{n(n-1)}{2} & = \nu_2(k!)\\ & = \sum _{i=1}^\infty \lfloor{\frac{k}{2^i}}\rfloor\\ & \le k \end{align*}Thus, $k \ge \frac{n(n-1)}{2}$. But then, note that $3\mid 2^i-1$ if and only if $2\mid i$. Further, by LTE we have for all even $i$, $\nu_3(2^i-1)= \nu_3(4-1) + \nu_3(\frac{i}{2}) = \nu_3(\frac{i}{2}) +1$. Thus, \begin{align*} \nu_3((2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1})) & = \nu_3(\lfloor \frac{n}{2} \rfloor) + \nu_3(\lfloor \frac{n-2}{2}\rfloor) + \dots + \nu_3(1)\\ &= \nu_3((\lfloor \frac{n}{2} \rfloor)!) + \lfloor \frac{n}{2} \rfloor \end{align*}Thus, as a result of our previous bound we have, \begin{align*} \nu_3((\lfloor \frac{n}{2} \rfloor)!) + \lfloor \frac{n}{2} \rfloor & = \nu_3(k!)\\ & \ge \nu_3((\lfloor \frac{n(n-1)}{2} \rfloor)!)\\ & \ge (n-1)\nu_3((\lfloor \frac{n}{2} \rfloor)!) \end{align*}which implies, \[\frac{n}{6}-1<\lfloor \frac{n}{6} \rfloor\le \nu_3((\lfloor \frac{n}{2} \rfloor)!) \le \frac{\lfloor \frac{n}{2} \rfloor}{n-2}\]from which it is not hard to see that we require $n<10$. Checking the possibilities $n<5$ we see that $n=1$ and $n=2$ work yielding $k=1$ and $k=3$ respectively and others don't ($n=4$ almost works). When $n\ge 5$, we have $31 \mid k!$. Thus, we must have $29 \mid k!$ as well. So, 29 is a fraction of some expression of the form $2^i-1$ which is clearly impossible for any $i<10$ which completes the check.

Let $s_{p}(n)$ be the sum of digits in the base $p$ expansion of $n$ , and the RHS be $Q$. Clearly \[ \nu_2(Q) = \frac{n(n-1)}{2}\]That shows $k \in \{2^n,2^n + 1\}$. Then observe that \[ \nu_p(k!) \leq \lfloor \frac{2^n}{p-1}\rfloor \leq \nu_p(Q)\]Clearly for all primes the $+1$ makes no diffrence. now the equality should hold for that $Q$ should only contain prime factors less than 7. as we want $\lfloor \frac{2^n}{p-1}\rfloor$ to be an integer. so 7 can't be in and hence no number above it. Manually checking for $k \in \{1,2,3,4,5,6\}$ we see that only $(1,1),(2,3)$ work

The only solutions are $(n, k) = (1, 1)$ and $(2, 3)$. The main idea is that for large enough $n$ (which will actually be pretty small), the right side has too many factors of $2$ to balance out the other prime factors. Specifically, in addition to examining $\nu_2$ we'll also look at $\nu_3$. Notice that \[\nu_2(k!) = \sum_{i = 0}^{n - 1} \nu_2(2^n - 2^i) = 1 + 2 + \dots + (n - 1) = \frac{n(n - 1)}{2}\]and \[\nu_3(k!) = \nu_3\left(\prod_{i = 0}^{n - 1} (2^n - 2^i)\right) = \sum_{i = 1}^n \nu_3(2^i - 1) = \sum_{i = 1}^{\lfloor n/2 \rfloor} \nu_3(4^i - 1) = \left\lfloor \frac n2 \right\rfloor + \sum_{i = 1}^{\lfloor n/2 \rfloor} \nu_3(i)\]\[ = \left\lfloor \frac n2 \right\rfloor + \nu_3\left(\left\lfloor \frac n2 \right\rfloor!\right)\]by LTE and the fact that $\nu_3(2^i - 1)$ is $0$ when $i$ is odd. This implies that $k > \lfloor n/2 \rfloor$, so \[\nu_3\left(\left(\left\lfloor \frac n2 \right\rfloor + 1\right) \cdot \left(\left\lfloor \frac n2 \right\rfloor + 2\right) \cdots k\right) = \left\lfloor \frac n2 \right\rfloor.\]At least $\left\lfloor \frac{k - \lfloor n/2 \rfloor}{3} \right\rfloor$ terms in the product on the left side are divisible by $3$, so we discover that \[\left\lfloor \frac n2 \right\rfloor \geq \left\lfloor \frac{k - \lfloor n/2 \rfloor}{3} \right\rfloor \geq \frac{k - \lfloor n/2 \rfloor}{3} - 1 \implies k \leq 4 \left\lfloor \frac n2 \right\rfloor + 3 \leq 2n + 3.\]But from before we have \[\frac{n(n - 1)}{2} = \nu_2(k!) = k - s_2(k) < k,\]so $\frac{n(n - 1)}{2} < 2n + 3$. Rearranging yields $(n - 6)(n + 1) < 0$, so $n < 6$. Checking these cases reveals that only $n = 1$ and $n = 2$ work as desired. $\square$

We'll show that if $n\geq 8$ then no such $k$ exists. First we'll prove a lemma. Lemma: If $n\geq 8$ then $\prod_{i=1}^n \left (2^n-2^{n-i}\right)>(2n)!$. Proof. Notice that $2^i-1>2i(2i-1)$ if $i\geq 8$ and we can check that $2^{8-i} \left (2^i-1\right )>2i(2i-1)$ if $i<8$. So we have $$\prod_{i=1}^n \left (2^{n-i}(2^i-1)\right)>\prod_{i=1}^n(2i(2i-1)),$$which is equivalent to the lemma. //// From the lemma, it follows that $k>2n$. So $$v_3(k!)\geq v_3((2n)!)=\sum_{i=1}^\infty \left\lfloor \frac{2n}{3^i} \right\rfloor \qquad (1)$$Now we state another lemma. Lemma: $2$ is a primitive root mod $3^m$ for $m=1,2,\dots$. Proof. Well known. Use LTE. //// By the lemma we have $2^i-1\equiv 0\pmod{3^m}$ iff $2\cdot 3^{m-1}=\phi(3^m)\mid i$. So, $$v_3\left (\prod_{i=1}^n \left (2^n-2^{n-i}\right)\right)=\sum_{i=1}^n v_3\left (2^i-1\right)=\sum_{i=1}^\infty \left \lfloor \frac{n}{2\cdot 3^{i-1}}\right \rfloor \qquad (2)$$Since $\frac{2n}{3^i}>\frac{n}{2\cdot 3^{i-1}}$ it follows that $\left \lfloor \frac{2n}{3^i}\right \rfloor \geq \left \lfloor \frac{n}{2\cdot 3^{i-1}}\right \rfloor$. But for $n\geq 8$ we have $$\left \lfloor \frac{2n}{3}\right \rfloor >\frac 23 n-1>\frac n2\geq \left \lfloor \frac{n}{2}\right \rfloor.$$So from $(1)$ and $(2)$ it follows that $$v_3(k!)>v_3\left (\prod_{i=1}^n \left (2^n-2^{n-i}\right)\right) \implies k!\neq \prod_{i=1}^n \left (2^n-2^{n-i}\right),$$as claimed. Now for $n<8$ we can manually check and find that only $n=1,2$ work.

I claim the answers are $(1,1), (2,3)$ only, it is trivial to verify that both of these work. To eliminate $n = 3,4$, compute the right hand side as $7 \cdot 6 \cdot 4, 7 \cdot 6 \cdot 4 \cdot 8 \cdot 15$ which is just $168, 168 \cdot 120$, the latter of which is just $4 \cdot 7!$, so neither of these are factorials. We take $\nu_2$ of both sides. We get $k \ge \nu_2(k!) = \nu_2(\prod 2^n - 2^i) = \frac 12 (n^2 -n)$. Now we can bound $\prod 2^n - 2^i < 2^{n(n - 1)} = 4^(\frac 12 (n)(n - 1))$. We claim that for $n > 4$ we have $ (\frac 12 (n^2 - n))! > 4^{\frac 12 (n^2 - n)}$, resulting in no solutions. We can rewrite this as $\prod_{1 \le i \le \binom n2} \frac i4 > 1$, the inequality $\frac i4 > 1$ is true for most $i$, to deal with the ones where its not we can just combine them with the highest values of $i$, thus we show for $x \ge 10$ the inequality $x(x - 1)(x - 2)\cdot (3 \cdot 2 \cdot 1) > 4^6$ is true, which is just combining the three highest terms with the three lowest. This is just obviously true by computation and the fact that the function is increasing.

Observe that \[\frac{n(n-1)}2=\nu_2\left(\prod_{i=0}^{n-1}(2^n-2^i)\right)=\nu_2(k!)\le k\]where the last step follows from evaluating Legendre's formula as a geometric series. Observe also that \[\frac{k}3-1<\nu_3(k!)=\nu_3\left(\prod_{i=0}^{n-1}(2^n-2^i)\right)=\nu_3\left(\prod_{i=1}^{n}(2^i-1)\right).\]Now $\nu_3(2^m-1)=0$ if $m$ is odd and $\nu_3(2^m-1)=\nu_3(4^{\frac{m}2}-1)=\nu_3(m)+1$ if $m$ by lifting exponents. Continuing from above, we find that \[\frac{k}3-1<\nu_3\left(\prod_{i=1}^{n}(2^i-1)\right)\le\frac{n}2+\sum_{i=1}^{\left\lfloor\frac{n}2\right\rfloor}\nu_3(2i)\le\frac{n}2+\nu_3(n!)<n\]where the last step again follows from evaluating Legendre's formula as a geometric series. Combining the above results imply that $n^2-n\le 2k<6n+6$, so $n<7$. Manually checking yields that the only solutions are $(k,n)=(1,1)$ and $(k,n)=(2,3)$. $\blacksquare$