$(n!)_p | (p-1)\times (p^2-1) ... (p^n-1)$ for any $p$ prime number?

Question

We note $(n!)_p$ is the simplified factorial by $p$,

it is the largest integer, prime with $p$ and which divides $n!$.

Is it true : for any $n\in \mathbb N$ such that $n> 1$ and $p$ prime integer : $$(n!)_p | (p-1)\times (p^2-1) ... (p^n-1)$$

Source : les dattes à Dattier

Answer 1

Claim: For all primes $p$ and non-negative integers $n$, $n!$ divides $$M=(p^{n}-p^{n-1})(p^n-p^{n-2})\cdots (p^n-p)(p^n-1).$$

Proof: (Outline) $M$ counts the number of ordered bases of a vector space of dimension $n$ over a field of $p$ elements. And for each unordered basis, there are exactly $n!$ different ordered bases.

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Now: $$M=p^{n(n-1)/2}(p-1)(p^2-1)\cdots (p^n-1)$$

If $D\mid M$ and $\gcd(D,p)=1$, then $D\mid (p-1)(p^2-1)\cdots(p^n-1)$.