You may just as well consider the remainders of $ \binom{2^n-1}{2k}$ for $ k=0,1,2,\dots,2^{n-1}-1$, and now, Hint 1:

Hint 2:

induction on $ n$ and we prove the srtonger assertion: $ \binom{2^{n}-1}{i}-\binom{2^{n}-1}{j}$ is not a multiple of $ 2^n$ for every $ (i,j)$ satisfies$ 1\le i<j \le 2^{n-1}-1$ and $ \binom{2^{n}-1}{i}+\binom{2^{n}-1}{j}$ is not a multiple of $ 2^{n+1}$ for every $ (i,j)$ satisfies$ 1\le i<j \le 2^{n-1}-1$ And I will post the full solution below(very sorry for being written in Chinese,I am glad to explain if anyone has questions)

I am grateful if anyone can point out my mistakes or flaws

I solved it by proving $ 1\le a\le b\le 2^{n-1}-1 \implies$ $ \prod_{x=a}^b \frac{2^n-x}{x}$ is not $ \equiv 1\pmod{2^n}$ and not $ \equiv -1\pmod{2^{n+1}}$ which is almost trivial by induction.

Let us note that $\binom{2^{n}-1}{k} = \prod_{j = 1}^{k}\frac{2^n-j}{j}$. Now, let us define the function F(n) = $\{ k \in \mathbb{Z} | 2^k | n, 2^{k+1} \nmid n\}$. We can see that F($2^n$-j) = F(j). From this we can see that $\binom{2^{n}-1}{k} \equiv 1$ mod $2$ $\forall$ $0 \le k \le 2^{n-1}-1$. We can also see that $F(j) \le 2^{n-2}$ since $j \le 2^{n-1}-1$. From this we can see that $\frac{2^n-j}{j} \equiv 3$ mod $4$. Now we notice that $\binom{2^n-2j}{2j} \equiv 1$ mod $4$ and $\binom{2^n-2j-1}{2j+1} \equiv 3$ mod $4$. Thus we now need only to show that $\binom{2^n-2j}{2j} \not\equiv \binom{2^n-2t}{2t}$ mod $2^n $ and $\binom{2^n-2j-1}{2j+1} \not\equiv \binom{2^n-2t-1}{2t+1}$ mod $2^n$ $\forall$ $j \neq t, 0 \le j, t \le 2^{n-2}-1$. To prove this we need only to show that $\prod_{j = a}^{b}\frac{2^n-j}{j} \not\equiv 1$ mod $2^n$ $\forall a \not\equiv b$ mod $2$. However, from Vieta's relatoins we can see that $\prod_{j = a}^{b}\frac{2^n-j}{j}$ can be rewritten as $\frac{2^n(2^n(x)+\sum_{j = a}^bj)}{\prod_{j = a}^bj}+1$ for some integer x. So we know want to show that $\frac{2^n(2^n(x)+\sum_{j = a}^bj)}{\prod_{j = a}^bj} \equiv 0$ mod $2^n$. However, we notice that $2^n(x)+\sum_{j = a}^bj$ is odd which means that there cannot be any factors of 2 in the denominator for $\frac{2^n(2^n(x)+\sum_{j = a}^bj)}{\prod_{j = a}^bj} $ to be divisible by $2^n$ and this is contradiction. Thus, the problem is proved.

A different approach: as $\binom{2^{n}-1}{i}$ is clearly odd, we need to show that $\binom{2^{n}-1}{j}/\binom{2^{n}-1}{i}$ is not 1 modulo $2^{n}$, if $1\le i<j\le 2^{n-1}-1$. We can do this by writing this as the product of $\frac{2^{n}-k}{k}$ for $i+1\le k\le j$. Now, choose $k$ such that $v_{2}(k)=m$ is maximized: it is easy to check that $k$ is unique. From here, after canceling powers of 2, it is easy to see that each term except than the one indexed by $k$ is $-1$ modulo $2^{n-m+1}$, and it is also easy to check that the exceptional one is not $\pm1$. Thus $\binom{2^{n}-1}{j}/\binom{2^{n}-1}{i}$ is not 1 modulo $2^{n-m+1}$, so if $m>0$ certainly it is not 1 modulo $2^{n}$ either. If $m=0$, on the other hand, we only have one factor in the product, so the fraction is equal to $-1$ modulo $2^{n-m+1}$, so we are done in this case as well.

April wrote: Let $ n$ be a positive integer. Show that the numbers \[ \binom{2^n - 1}{0},\; \binom{2^n - 1}{1},\; \binom{2^n - 1}{2},\; \ldots,\; \binom{2^n - 1}{2^{n - 1} - 1}\] are congruent modulo $ 2^n$ to $ 1$, $ 3$, $ 5$, $ \ldots$, $ 2^n - 1$ in some order. Proposed by Duskan Dukic, Serbia I finally solved it , after trying it since days! . It is equivalent to show that $\binom{2^n-1}{1},\binom{2^n-1}{3}, \cdots , \binom{2^n-1}{2^{n}-1}$ leaves distinct remainders . Suppose for the contrary that $\binom{2^n-1}{x} \equiv \binom{2^n-1}{y}\pmod{2^n}$. We can take $x>y$ . Using the binomial identities we get, $\binom{2^n}{x}-\binom{2^n}{x-1}+\binom{2^n}{x-2}+\cdots-\binom{2^n}{y+1} \equiv 0 \pmod{2^n}$. Now , since there are powers of $2$ , it is handy to use kummer's theorem . we will find $v_2{\binom{2^n}{k}}$ . Suppose that $v_2(k)=q$ then using kummer';s theorem we get $v_2{\binom{2^n}{k}} = n-q$ . Note that , if k is odd then $q=0$. Thus , in this case $2^n$ will divide $\binom{2^n}{k}$ . So we have to only check the sum $k$ which are even. Also we note that if $k$ is even then $v_2{\binom{2^n}{k}} < n$ . Now , we choose a $k=a$ such that $v_2{\binom{2^n}{a}}=w$ is minimal. Then we will be able to take $2^w$ common , and the left term will be odd . so there will be a term where $k=b$ such that , $v_2{\binom{2^n}{a}} = v_2{\binom{2^n}{b}}$ . Now from earlier observation we see that , more the $v_2{k}$ less will be $v_2{\binom{2^n}{k}}$ . here we have , $v_2{\binom{2^n}{a}} = v_2{\binom{2^n}{b}} \implies v_2(a)=v_2(b)$. Suppose that $a<b$. Obviously $\exists c$ between $a$ and $b$ such that $v_2(a)<v_2(c)$ . Contradicting the minimality of $v_2{\binom{2^n}{x}}$ . So we are done! $\Box$ ^^ Someone please check it , i am not sure it it is correct.

WOuld you please explain what's $v_2{\binom{2^n}{k}}$ ?? I'm dying to understand this, especially kummer's theorem !??

Hmm, I'd like to point out a solution without using valuations, for those who don't know it... All such numbers are odd, which is easy to check. Lemma 1: $\dbinom{2^n - 1}{2k} + \dbinom{2^n-1}{2k+1} \equiv 2^n \pmod{2^{n+1}}$ Proof: Observe that $\dbinom{2^n - 1}{2k}(1 + \frac{2^n - (2k+1)}{2k+1}) = \dbinom{2^n - 1}{2k}(\frac{2^n}{2k+1}) \equiv 0 \pmod{2^n}$, but it doesn't divide $2^{n+1}$, so it is equivalent to $2^n \pmod{2^{n+1}}$ Claim: $\dbinom{2^n - 1}{4k+3} \equiv \dbinom{2^{n-1}-1}{2k+1} \pmod{2^n}$ for defined values of $k$ Proof: We use induction. Check the result for $k = 0$. Now observe that for $k \rightarrow k+1$, it suffices to show that $\frac{\dbinom{2^n - 1}{4k+7}}{\dbinom{2^n - 1}{4k + 3}} \equiv \frac{\dbinom{2^{n-1} - 1}{2k+3}}{\dbinom{2^{n-1} - 1}{2k+1}} \pmod{2^n}$ Observe that the first product is equivalent to $(\frac{2^n - (4k+7)}{4k+7})(\frac{2^n - (4k+6)}{4k+6})(\frac{2^n - (4k+5)}{4k+5})(\frac{2^n - (4k+4)}{4k+4}) \equiv (-1)^2 (\frac{2^{n-1} - (2k+3)}{2k+3})(\frac{2^{n-1} - (2k+2)}{2k+2}) \equiv \frac{\dbinom{2^{n-1}-1}{2k+3}}{\dbinom{2^{n-1} - 1}{2k+1}} \pmod{2^n} $ Claim: $\dbinom{2^n - 1}{4k} \equiv \dbinom{2^{n-1} - 1}{2k} \pmod{2^n}$: Proof: Similar to above. Denote $P_n$ the assertion for $n$. We use induction. The result is true for $n = 1, 2, 3$, and $4$. We prove it for larger values. So for induction, given $P_n$, observe that each term in $S_n$, the values of the binomial coefficients, are distinct mod $2^n$. When we use the above relations, we obtain a $P_{n+1}$, after using the lemma. How could there be a duplicate in this set? Well, if there is one, it would have to be a number from $S_n$, and then the lemma would produce another value. These two values would somehow be the same if the duplicate occurred. We show that this cannot happen. Observe that this implies that in $S_n$, there are two numbers such that, when summed, give $0 \pmod{2^{n+1}}$. First, the sum is obviously $0 \pmod{2^n}$, which each element in $S_n$ has only one pair for: one of its neighbors. But the sum, as shown, isn't $0 \pmod{2^{n+1}}$, so we are done.

April wrote: Proposed by Duskan Dukic, Serbia Actually his name is Dušan Djukić, or if you don’t have Serbian latin letters, Dusan Djukic is fine.

Induct, with base cases trivial. Our inductive hypothesis implies that there exist precisely two solutions $x=m$ and $x=2^{n-1}-1-m$ satisfying $\binom{2^{n-1} - 1}{x} \equiv L \pmod{2^{n-1}}$ for any given odd number $L$. Note that \[\binom{2^n-1}{2k} = \prod_{i=1}^{2k} \frac{2^n-i}{i} = \left( \prod_{i=1}^{k} \frac{2^{n-1}-i}{i} \right) \left(\prod_{\substack{j=1, \\ j \text{ odd}}}^{2k-1} \frac{2^n-j}{j}\right) = \binom{2^{n-1}-1}{k} \left(\prod_{\substack{j=1, \\ j \text{ odd}}}^{2k-1} \frac{2^n-j}{j}\right).\]We can safely take the left and right ends of this equality modulo $2^n$, whence $\binom{2^n-1}{2k} \equiv (-1)^k\binom{2^{n-1}-1}{k} \pmod{2^n}$. Now suppose that $\binom{2^n-1}{2i} \equiv \binom{2^n-1}{2j} \pmod {2^n}$ so that \[(-1)^i\binom{2^{n-1}-1}{i} \equiv (-1)^j\binom{2^{n-1}-1}{j} \pmod{2^n}.\]Some casework: If $i$ and $j$ are of the same parity, then the above binomial coefficients are also congruent modulo $2^{n-1}$, so by hypothesis either $i=j$ or $i=2^{n-1} -1- j$. Parity forces $i=j$. If $i$ and $j$ are of different parity, we have \[\binom{2^{n-1}-1}{i} \equiv -\binom{2^{n-1}-1}{j} \pmod{2^n}. \quad (\heartsuit)\]WLOG $i$ is even. This congruence $(\heartsuit)$ holds modulo $2^{n-1}$, but we can also check that \[\binom{2^{n-1}-1}{i} \equiv -\frac{2^{n-1}-(i+1)}{i+1} \binom{2^{n-1}-1}{i} = -\binom{2^{n-1}-1}{i+1} \pmod{2^{n-1}}\]\[\implies \binom{2^{n-1}-1}{j} \equiv \binom{2^{n-1}-1}{i+1} \pmod{2^{n-1}}.\]By hypothesis, this forces either $j=i+1$ or $j=2^{n-1}-1-(i+1)$, but $i$ and $j$ have different parity so we must have $j=i+1$. Now examine $(\heartsuit)$. Rewrite the right-side binomial coefficient to get \[\binom{2^{n-1}-1}{i} \equiv -\frac{2^{n-1}-(i+1)}{i+1} \binom{2^{n-1}-1}{i} \pmod{2^n} \]\[\implies \frac{2^{n-1}-(i+1)}{i+1} \equiv -1 \implies 2^{n-1} \equiv 0 \pmod{2^n},\]which is absurd. So this case is impossible. The above work shows that $\binom{2^n-1}{2k}$ for $k=0, 1, \dots, 2^{n-1}-1$ are distinct odd numbers modulo $2^n$, which easily converts to the desired claim.

Our solution proceeds in two parts. First we show that $\binom{2^n-1}{a}$ is odd for any integer $a$. Then we show that \[ \binom{2^n-1}{a} \not\equiv \binom{2^n-1}{b} \pmod{2^n} \]for any $0 \leq a < b \leq 2^{n-1}-1$. It is easy to see that these two claims put together give the result. The first part is rather simple. By Legendre, \[ \nu_2 \left( \binom{2^n-1}{a} \right) = s_2(a) + s_2(2^{n-1}-1-a) - s_2(2^n-1) \]Since no carries occur, this is just zero. Most of the problem is in the second part. It suffices to show that \[ \prod_{k=b+1}^a \frac{2^n - k}{k} \not\equiv 1 \pmod{2^n} .\]Let $M = \underset{b < k \leq a}{\max} \nu_2(k)$. Denote by $t$ the unique integer in that range where the maximum is taken.Clearly, $M \leq n-2$. Let $t = 2^M m$. Then \begin{align*} \prod_{k=b+1}^a \frac{2^n - k}{k} &= \prod_{k=b+1}^a \frac{2^{n- \nu_2(k)} - k/\nu_2(k)}{k /\nu_2(k} \\ & \equiv (-1)^{a-b-1} \frac{2^{n-M} - m}{m} \\ & \equiv \pm (2^{n-M} - 1) . \pmod{2^{n-M+1}} . \end{align*}At the last step, we make use of the fact that $m^{-1}$ is odd. Since $2^{n-M} \geq 2$, it is impossible that \[ 2^{n-M} - 1 \equiv \pm 1 \pmod{2^{n-M + 1}}. \]Since \[ \binom{2^n-1}{a} = \binom{2^n-1}{b} \prod_{k=b+1}^a \frac{2^n - k}{k} \]we are done.

Before starting with the proof, we establish that all the desired binomial coefficients are odd, which allows more ease in manipulating them modulo $2^n$. Claim: We have that $\binom{2^n-1}{k}$ is odd for $0\le k\le 2^n-1$. Proof: Let $s_2(m)$ denote the sum of the digits of $m$ written in binary. It is well known that \[v_2\left(\binom{a+b}{a}\right) = s_2(a+b)-s_2(a)-s_2(b).\]Using this, we see that \[v_2\left(\binom{2^n-1}{k}\right) = s_2(2^n-1) - s_2(k) - s_2((2^n-1)-k).\]Since $2^n-1$ is $n$ ones written in binary and $0\le k\le 2^n-1$, we see that the set of ones in the binary representation of $2^n-1-k$ is exactly the complement of those in the binary representation for $k$, so in fact \[s_2(2^n-1) - s_2(k) - s_2((2^n-1)-k)=0.\]This proves the claim. $\blacksquare$ We proceed by induction on $n$, with $n=1$ trivial. Now suppose the problem is true for $n$. Before proving $n+1$, we need to prove the following intermediate claim. Claim: The numbers \[(-1)^a\binom{2^n-1}{a}\]generate all the odd residues modulo $2^{n+1}$ as $a$ varies from $0$ to $2^n-1$. Proof: It suffices to show that \[(-1)^{a_1}\binom{2^n-1}{a_1} \equiv (-1)^{a_2}\binom{2^n-1}{a_2}\pmod{2^{n+1}}\]implies that $a_1=a_2$. The inductive hypothesis implies that in fact, $\binom{2^n-1}{2b}$ generates all the odd residues mod $2^n$ as $b$ varies from $0$ to $2^{n-1}-1$, and a similar statement holds if we replace $2b$ with $2b+1$. Thus, if $a_1$ and $a_2$ have the same parity, we are done trivially by simply looking modulo $2^n$. So WLOG suppose that $a_1$ is even and $a_2$ is odd. Note that \[\binom{2^n-1}{a_2}=\frac{2^n-a_2}{a_2}\binom{2^n-1}{a_2-1},\]and that \[\frac{2^n-a_2}{a_2} = \frac{1}{a_2}2^n - 1 \equiv 2^n-1\pmod{2^{n+1}},\]so \[\binom{2^n-1}{a_2}\equiv (2^n-1) \binom{2^n-1}{a_2-1}\pmod{2^{n+1}}.\]Taking our original equation mod $2^n$ combined with the above implies that \[\binom{2^n-1}{a_1}\equiv \binom{2^n-1}{a_2-1}\pmod{2^n},\]and since both $a_1$ and $a_2-1$ are even, the inductive hypothesis implies that $a_1=a_2-1$. However, this them implies \[\binom{2^n-1}{a_2}\equiv (1-2^n)\binom{2^n-1}{a_2-1}\pmod{2^{n+1}},\]which is a clear contradiction as we can cancel out the binomial coefficients since they are odd. This is the desired contradiction, so $a_1$ and $a_2$ must in fact have the same parity, which we already resolved. This completes the proof. $\blacksquare$ The induction is now fairly clear. We see that \[\binom{2^{n+1}-1}{2a} = \frac{\prod_{i=0}^{a-1}(2^{n+1}-1-2i)}{\prod_{i=0}^{a-1}(1+2i)}\binom{2^n-1}{a}\equiv (-1)^a\binom{2^n}{a}\pmod{2^{n+1}},\]so \[\binom{2^{n+1}-1}{2a}\]generates all odd residues mod $2^{n+1}$, which completes the induction, and thus the proof.

. This solution is almost the same as #15 and in my opinion, is cleaner than the 'inductive/recursive' solutions above. The fact that each of the binomial coefficients is odd is obvious because $\binom{2^n-1}{k} = \prod_{i=1}^{k} \frac{2^n-i}{i}$ where $i \leq 2^{n-1}-1$ hence $\nu_2(i) \leq n-1$ meaning that $\nu_2(2^n-i) = \nu_2(i)$ which means that $$\nu_2( \frac{2^n-i}{i}) = \nu_2(2^n-i)-\nu_2(i) = 0$$and therefore $\binom{2^n-1}{k}$ is odd for all $k \in \{0, \cdots, 2^{n-1}-1\}$. We will now prove that each of these binomial coefficients is distinct in $\pmod{2^n}$, assume for the sake of contradiction otherwise. Then there exists $a>b$ among $\{0, \cdots 2^{n-1}-1\}$ such that $\binom{2^n-1}{a} \equiv \binom{2^n-1}{b} \pmod{2^n}$ Therefore, $$\prod_{i=b+1}^{a} \binom{2^n-i}{i} \equiv 1 \pmod{2^n}$$as odds have inverses.

. $\textbf{Lemma 1:}$ For any set of consecutive positive integers, there exists an element with maximal $\nu_2()$. $\textbf{Proof)}$ The proof is almost trivial, assume for the sake of contradiction that this is not the case, that is the maximal power of two is achieved twice, say at $u$ and $v$, then $u = 2^M \cdot k, v = 2^M \cdot t$ for odd positive integers $k,t$ such that without loss of generality $k < t$, then as $k \equiv t \equiv 1 \pmod{2}$, $k+1 < t$ as well and hence $s = 2^M(k+1)$ is among the integers achieving a higher $\nu_2()$, as desired. $\blacksquare$

Now, we can use $\textbf{Lemma 1}$ and consider the positive integer with maximal $\nu_2(t) = m$ among $b+1, \cdots, a$ for which $m \leq n-2$ because we are only considering $\binom{2^n-1}{0}, \binom{2^n-1}{1}, \cdots, \binom{2^n-1}{2^{n-1}-1}$. Then all but one of the fractions are of the form $\frac{2^T-s}{s}$ for odd $s$ and $T > n-m$ meaning that $$(-1)^{a-b-1} \cdot \frac{2^{n-m}-s}{s} \equiv \prod_{i=b+1}^{a} \binom{2^n-i}{i} \equiv 1 \pmod{2^{n-m+1}}$$for some $s$ odd which therefore means that $$(-1)^{a-b-1} \cdot s+s \equiv 2^{n-m} \pmod{2^{n-m+1}}$$as $2^{n-m} \not\equiv 0 \pmod{2^{n-m+1}}$ we must have that $$2s \equiv 2^{n-m} \pmod{2^{n-m+1}}$$which because $s$ is odd implies that $n-m = 1$ contradicting the fact that $m \leq n-2$, as desired. $\blacksquare$

Fairly sure this works. The proof proceeds in two parts. Part I: Parity. I claim that ${2^n-1 \choose i}$ is odd for all $0 \leq i \leq 2^n-1$. This follows by Lucas' Theorem because $2^n-1 = \overline{111\cdots 1}_2$, and thus subtracting any integer from it does not induce any carrying in base two. Part II: Uniqueness. It remains to show that $${2^n-1 \choose i} \not \equiv {2^n-1 \choose j} \pmod {2^n}$$for $0 \leq i \neq j \leq 2^{n-1}-1$. Divide both sides of the congruence by ${2^n-1 \choose i}$, which is valid as the number is odd: $$\frac{{2^n-1 \choose j}}{{2^n - 1 \choose i}} \equiv \frac{(2^n-i-1)(2^n-i-2)\cdots(2^n-j)}{(i+1)(i+2)\cdots (j-1)j} \pmod {2^n}.$$We induct on $n$, with the stronger hypothesis that this expression cannot be congruent to 1 modulo $2^n$, and furthermore that it cannot be congruent to $-1$ modulo $2^n$ unless $|i-j| = 1$. The base case $n=2$ can be checked manually, and furthermore, notice that any two consecutive terms $$\frac{2^n - i}i \equiv -1 \pmod {2^n}$$if $i$ is odd, and we may reduce to the $2^{n-1}$ case if $n$ is even, so the $|i-j| = 1$ base case holds. Next, suppose for the sake of contradiction that there exist $i < j$ such that the expression is congruent to $-1$ or 1 modulo $2^n$. Because $\nu_2(2^n-i-1) = \nu_2(i+1)$ for all $i$, we may remove terms on the numerator and denominator that are both odd and cancel out a factor of 2 from the remaining terms: $$(-1)^a \equiv \frac{(2^n-i-1)(2^n-i-2)\cdots(2^n-j)}{(i+1)(i+2)\cdots (j-1)j} \equiv (-1)^b \prod_{k=i'}^{j'} \frac{2^{n-1} - k}k \pmod {2^n}$$for some positive integers $i', j', a, b$. Notice that $j' \leq \frac j2 \leq 2^{n-2} - 1$ and obviously $i' \geq 0$, so we may apply the inductive hypothesis to obtain $$\prod_{k=i'}^{j'} \frac{2^{n-1} - k}k \not \equiv (-1)^{r_{n-1}} \pmod {2^{n-1}} \implies \prod_{k=i'}^{j'} \frac{2^{n-1} - k}k \not \equiv (-1)^{r_{n-1}} \pmod {2^n}.$$If there is only one term in the reduced expansion that is congruent to $-1 \pmod {2^{n-1}}$, notice that $$\frac{2^{n-1} - i}i \not \equiv -1 \pmod {2^n},$$so the quotient cannot be congruent to $-1$ either. This yields a contradiction, so the inductive step is complete. As a result, all the binomial coefficients are congruent to distinct odd residues mod $2^n$. It follows they must be congruent to $1, 3, 5, \cdots, 2^n - 1$ in some order, as required.

Can someone check this? For some reason I'm terrible at algebra and this took me two hours. The fact that they are all odd is evident by $\nu_2(i)<n-1\Rightarrow\nu_2(2^n-i)=\nu_2(i)$, so $\tbinom{2^n - 1}{2k}(1 + \tfrac{2^n - (2k+1)}{2k+1}) = \tbinom{2^n - 1}{2k}(\tfrac{2^n}{2k+1}) \equiv 2^n \pmod{2^{n+1}}$. Now by induction we prove that $\tbinom{2^n - 1}{4k+3} \equiv \tbinom{2^{n-1}-1}{2k+1} \pmod{2^n}$, base case evident, with inductive step following $$(\tfrac{2^n - (4k+7)}{4k+7})(\tfrac{2^n - (4k+6)}{4k+6})(\tfrac{2^n - (4k+5)}{4k+5})(\tfrac{2^n - (4k+4)}{4k+4}) \equiv (\tfrac{2^{n-1} - (2k+3)}{2k+3})(\tfrac{2^{n-1} - (2k+2)}{2k+2}) \equiv \tfrac{\binom{2^{n-1}-1}{2k+3}}{\binom{2^{n-1} - 1}{2k+1}} \pmod{2^n};$$similarly we deduce $\binom{2^n-1}{4k+1}\equiv\binom{2^{n-1}-1}{2k+1},\binom{2^n - 1}{4k} \equiv \binom{2^{n-1} - 1}{2k}\equiv\binom{2^n-1}{4k+2}\pmod{2^n}$. The problem is finished by inductive hypothesis of only odd residues appearing in tandem with noting that $\binom{2^{n+1}-1}{4k+1}\not\equiv\binom{2^{n+1}-1}{4k+3}\pmod{2^{n+1}}$, which follows from basic algebra.

Obviously, these are all odd by Lucas's Theorem. Now, we claim that they must all be distinct. Assume that, for some $i < j,$ we have $${2^{n}-1 \choose i} \equiv {2^{n}-1 \choose j} \pmod{2^{n}}.$$Then, \begin{align*} {2^{n}-1 \choose j} - {2^{n}-1 \choose i} & = \frac{(2^{n}-1)! ((2^{n}-1-i)!i! - (2^{n}-1-j)!j!)}{i! j! (2^{n}-1-i)! (2^{n}-1-j)!}\\ & = {2^{n}-1 \choose i} \cdot \frac{i!(2^{n}-1-j)! [(2^{n}-i-1) \cdots (2^{n}-j)-j \cdots (i+1)]}{j! (2^{n}-1-j)!} \\ & = {2^{n}-1 \choose i} \cdot \frac{i!}{j!} \cdot \left(\prod_{k = i+1}^{j} (2^{n}-k) - \prod_{k = i+1}^{j} k \right). \end{align*}Notice that $j-i$ is even, so we can rewrite the first product as $\prod_{k = i+1}^{j}(k-2^{n}).$ So, when considering the number of factors of $2,$ the expression is just $$\frac{i!}{j!} \cdot \sum_{k=1}^{j-i}2^{nk} S_{(j-i)-k}(i+1, \cdots, j).$$ Claim: $\frac{i!}{j!}S_{j-i-1}(i+1, \cdots, j)$ has a negative $2$-adic. Proof: Notice that $$\frac{i!}{j!} S_{j-i-1}(i+1, \cdots, j) = \sum_{k=i+1}^{j}\frac{1}{k}.$$Consider the largest $k$ such that there exists an $\ell$ with $i < \ell \le j,$ and $\nu_{2}(\ell) = k.$ We claim that only such one $\ell$ can exist. Considering the smallest such $\ell,$ if there are more possible values, then $\ell+2^{k+1}$ must work as well, as it clearly has $2$-adic equal to $k.$ But, then, if both of these numbers are in the range $[i+1,j],$ then so is $\ell+2^{k},$ which has $2$-adic at least $k+1,$ a contradiction. So, $$\sum_{k=i+1}^{j} \frac{1}{k} = \nu_{2}\left(\frac{1}{\ell}\right)=-k < 0,$$and thus we have proven the desired. $\blacksquare$ So, from the claim, $$\frac{i!}{j!} \cdot \sum_{k=1}^{j-i}2^{nk} S_{(j-i)-k}(i+1, \cdots, j) $$$$= \nu_{2}\left(2^{n} \cdot \frac{i!}{j!} \cdot S_{j-i-1}(i+1, \cdots, j)+2^{2n}\left(\sum_{k=2}^{n} 2^{(k-2)n}S_{(j-i)-k}(i+1, \cdots, j)\right)\right) < n,$$and so $2^{n} \nmid {2^{n}-1 \choose i} - {2^{n}-1 \choose j},$ as desired.