Look the prime between $\frac{n}{2}<p<n$

I feel I overcomplicate a lot but anyway. Let us present a solution. It can be manually checked that numbers $n=1,2,3$ hold the condition. Now we'll prove these are the only solutions. Let $n\geq 4$. Now take the prime $p$ which is bigger than $ \frac{n}{2}$ and smaller than $n$. This prime is obviously odd. Also, let $q$ be a prime divisor of $2^p-1$. From the order argument, it's open that $q>2p>n$. And from LTE $$v_q(2^{n!}-1)=v_q(2^p-1)+v_q(\frac{n!}{p})=v_q(2^p-1)$$which means the numbers $2^p-1$ and $\frac{2^{n!}}{2^p-1}$ are coprime. Also since $2^p-1$ and $2^n-1$ are coprime $2^p-1$ must be a perfect square. But this is impossible since $-1$ is not a $QR$ at $\pmod{4}$. Thus the only solutions to the given condition are $n=1,2,3$.

You can bypass the order argument. Suppose $n\ge 4$ and $p$ be a prime such that $n/2<p<n$. Note that $(2^p-1,2^n-1)=2^{(p,n)}-1=1$ and $2^p-1\mid 2^{n!}-1$. So, for some $x$, \[ x^2 = \left(2^p-1\right)\cdot \frac{2^{n!}-1}{(2^n-1)(2^p-1)}, \]where the second term above is integral as $(2^p-1,2^n-1)=1$. Since the terms in the product are coprime, it holds that $2^p-1$ is a square, which is impossible as $2^p-1\equiv 3\pmod{4}$.

Great problem! A bit scary for P1 in the day, but I think this simple argument works. Direct verification shows that $n=1,2,3$ work (the ratios are $1$, $1$, $3^2$) and $n=4$ does not (the ratio is $2^{18} + 2^{12} + 2^6 + 1 = (2^6 + 1)(2^{12}+1)$ and is divisible by $5$ but not by $25$), let $n\geq 5$. To make this less confusing, write for convenience $a = 2^n$ (which is even) and $t = (n-1)!$ (which is a multiple of $4$ when $n\geq 5$), so we work with $\frac{a^t-1}{a-1} = a^{t-1} + a^{t-2} + \cdots + a + 1 = \left(a^{\frac{t}{2}} + 1\right)\left(a^{\frac{t}{2} - 1} + a^{\frac{t}{2} - 2} + \cdots + a^2 + a + 1\right)$. Note that both factors are coprime, since a common divisor would divide both $a^{\frac{t}{2}} + 1$ and $a^{\frac{t}{2}} - 1$ and hence $2$, which is impossible since $a$ is even. Hence $a^{\frac{t}{2}} + 1 = x^2$ for some positive integer $x$, thus $(x - a^{\frac{t}{4}})(x + a^{\frac{t}{4}}) = 1$, impossible as both factors cannot be equal.

May be a scarry problem for p1 but it is not that hard my solution was same as @above When you see similar fractions you can use the same technique...