$n!$ divides a product: Part I

It is, because $S_n$ embeds in $GL_n({\mathbb F}_2)$.

Essentially the same argument: the ratio is a number of bases of the vector space $\mathbb{F}_2^n$.

A proof working also for composite values of $a=2$:

Let $p$ be a prime dividing $n!$. If $p$ divides $a$, then the product $\prod (a^n-a^k)$ is divisible by $p^{n-1}$, while $n!$ is not divisible by $p^n$. If $a$ and $p$ are coprime, it suffices to check that for any exponent $s$ the set $\{a^n-a^k\}$ contains at least as many numbers divisible by $p^s$ as the set $\{1,\dots,n\}$. It follows from the fact that a period of powers of a sequence $\{1,a,a^2,\dots\}$ modulo $p^s$ is less than $p^s$.