Grassmanian : arithmetic proof for its cardinality? Let $p$ be a prime and $n$ an integer greater that 1.
It is well known that :
$$\text{card}\left(GL_n(\mathbb{F}_p)\right)=\prod_{k=0}^{n-1}\left(p^n-p^k\right)$$
Since the n-th symmetric group $\mathfrak{S}_n$ can be seen as a subgroup of $GL_n(\mathbb{F}_p)$ (because every permutation matrix is invertible), we can deduce from Lagrange theorem that :
$$n!\mid\prod_{k=0}^{n-1}\left(p^n-p^k\right)$$
I am looking for some direct proof of this result, using only arithmetical arguments (divisibility, congruences, Gauss theorem, Bezout identity ...).
 A: Let $q \leq n$ be a prime number. What is the $q$-adic valuation of $P=\prod_{k=0}^{n-1}{p^n-p^k}$?
If $q \neq p$, then let $1 \leq \omega < n$ be the multiplicative order of $p$ mod $q$. Then the $q$-adic valuation of $P$ is the sum of the $q$-adic valuations of the $N_t=p^n-p^{n-t\omega}$ for $0 < t\omega \leq n$. But the $q$-adic valuation of $N_t$ is that of $p^{t\omega-1}$, which is (LTE lemma) the $q$-adic valuation of $t$ plus that of $p^{\omega}-1$.
In other words $v_q(N_t)=v_q(t)+v_q(p^{\omega}-1)$.
Let $\Omega$ be the integer part of $N/\omega$.
So $v_q(P)=\sum_{1 \leq t \leq \Omega}(v_q(t))+\Omega v_q(p^{\omega}-1)=\Omega v_q(p^{\omega}-1)+v_q(\Omega!)$.
If $\Omega \geq q$, then $v_q(P) \geq \Omega+1 \geq \frac{n}{\omega} \geq \frac{n}{q-1}\geq v_q(n!)$ by Legendre’s formula.
If not, then $q+1 > \frac{n}{\omega} \geq \frac{n}{q-1}$ so $n < q^2$ and by Legendre’s formula $v_q(n!)$ is the integer part of $n/q$, which is lower than or equal to $\Omega \leq v_q(P)$.
Finally, $v_p(P) \geq v_p(p^n-p^{n-1}) \geq n-1 \geq v_p(n!)$. So for any prime divisor $q$ of $n!$, $v_q(n!) \leq v_q(P)$, so $n!|P$.
