n! divides a product: Part II

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).