This is a follow up on another MO question.
Question. For $n\geq2$, the following is always an integer. Is it not? $$\frac{(2^n-2)(2^{n-1}-2)\cdots(2^3-2)(2^2-2)}{n!}.$$
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Sign up to join this communityThis is a follow up on another MO question.
Question. For $n\geq2$, the following is always an integer. Is it not? $$\frac{(2^n-2)(2^{n-1}-2)\cdots(2^3-2)(2^2-2)}{n!}.$$
As $S_n$ also embeds in $GL_{n-1}({\mathbb F}_2)$, you only need to check that $n!$ is not divisible by $2^n$.
Here is a direct number theoretic proof. First of all note that $$(2^n-2)(2^{n-1}-2)\cdots(2^2-2)=2^{n-1}(2^{n-1}-1)(2^{n-2}-1)\cdots(2^1-1).$$ Now for any prime $p$ we have, by Legendre's formula and by Fermat's little theorem, $$v_p(n!)\leq\left\lfloor\frac{n-1}{p-1}\right\rfloor\leq v_p\bigl(2^{n-1}(2^{n-1}-1)(2^{n-2}-1)\cdots(2^1-1)\bigr),$$ where for the second inequality we argue separately for $p=2$ and for $p\geq 3$. The result follows.
P.S. The above proof was inspired by Cherng-tiao Perng's deleted response (which was not entirely correct).