Prove that $\frac{(p^{n}-1)(p^{n}-p).....(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$

Solution 1:

I've got an idea how to solve this problem. The proof is not complete, but I am rather confident that one can complete it.

Let's show more generally that $k!$ divides $(p^n-1) \cdot \dotsc \cdot (p^n-p^{k-1})$. I believe that this is true for every integer $p$. Recall that the multiplicity of a prime number $q$ in $k!$ equals $$v_q(k!)=\sum_{\ell \geq 1} \bigl\lfloor \frac{k}{q^\ell} \bigr\rfloor.$$ Thus, it suffices to prove that the multiplicity of $q$ in the product $P=(p^n-1) \cdot \dotsc \cdot (p^n-p^{k-1})$ is at least that large. Now the idea is to decompose $P$ in partial products $P_\ell$ such that the multiplicity of $q$ in $P_\ell$ is at least $\lfloor \frac{k}{q^\ell} \rfloor$, so that in the end the multiplicity of $P$ is at least $\sum_\ell \lfloor \frac{k}{q^\ell} \rfloor$.

I can show at least that the multiplicity is at least $\lfloor \frac{k}{q} \rfloor$: Consider $P_i = (p^n-p^{i q}) \dotsc (p^n - p^{iq+q-1})$ for $i \leq \lfloor \frac{k}{q} \rfloor-1$, so that $\prod_i P_i$ divides $P$. Then $P_i \equiv 0 \bmod q$, since each power of $p$ is one the powers of $1,\dotsc,p^{q-1}$ modulo $q$. Hence, $q^{\lfloor \frac{k}{q} \rfloor}$ divides $P$.