Proof that given $p$ prime and $n \in \mathbb{N}$, $n!$ divides $(p^n - 1)(p^n - p)\cdots(p^n - p^{n-1})$.

Question

I was hoping to show this by Lagrange's theorem. As the general linear group over $\mathbb{F}_p$ (the field with $p$ elements), $\mathrm{GL}_n(\mathbb F_p)$ has order $(p^n - 1)(p^n - p)\cdots(p^n - p^{n-1})$, we just need to find a subgroup that has an order of $n!$.

We know that $S_n$ (symmetric group) has order $n!$ and since it is isomorphic to the set of permutation matrices with matrix multiplication as its operation, we already have or proof constructed.

My problem here is that it works for every $p \in \mathbb{N}$, it is not required for $p$ to be prime. Where am I wrong? Am I wrong?

Answer 1

It only works for $p$ prime, because otherwise you don't have a vector space over the field $\Bbb F_p$.