For $n\in\mathbb{Z}^{+}$, consider the polynomials
$$P\_n(x)=\prod\_{k=0}^{n-1}(x^n-x^k).$$



> 
> **QUESTION.** Is it possible to find a closed formula for the number of monomials in $P\_n(x)$, after expansion?
> 
> 
> 


Those interested in such enumeration may like to [read this](https://arxiv.org/pdf/0811.3652.pdf).


Here is an assertion which might be possible to prove.



> 
> **CLAIM.** For any $m\in\mathbb{Z}$, the following are all integers:
>  $$\frac1{n!}P\_n(m).$$
> 
> 
> 


**UPDATE.** If it helps, I've found a stronger divisibility when $m$ is a prime $p$.


$$\frac{P\_n(p)}{n!\,\cdot p^{\eta\_p(n)}}=\frac{p^{\binom{n}2}\prod\_{k=1}^n(p^k-1)}{n!\,\cdot p^{\eta\_p(n)}}$$
where $\eta\_p(n)=\nu\_p\left((p\lfloor n/p\rfloor)!\right)$ and $\nu\_p(m)$ is the [$p$-adic valuation](https://en.wikipedia.org/wiki/P-adic_order) of an integer $m$.



counting monomials and integrality
The claim is correct. We begin with a standard lemma.


**Lemma:** Let $p$ be an odd prime, and $x$ an integer coprime to $p$. Let $k$ be a positive integer and let $o$ be the multiplicative order of $x$ modulo $p$. Then $$v\_p(x^k-1) = \begin{cases} v\_p(x^o-1) + v\_p(\frac{k}{o}) & o \mid k \\ 0 & o \nmid k \end{cases}.$$


We prove the claim by comparing the [$p$-adic valuation](https://en.wikipedia.org/wiki/P-adic_order) of $P\_n(m)$ and $n!$ for every prime $p \le n$ (since all prime divisors of $n!$ are $\le n$). We skip the primes in the set $\{ q: q \mid m \} \cup \{2\}$ (they may be treated separately). The lemma implies that your claim is equivalent to the following inequalities:
$$(\*) \forall p\le n (\text{odd, coprime to }m): v\_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor v\_p(m^o-1) \ge v\_p(n!),$$
$$\text{where }o \text{ is the multiplicative order of m modulo p},$$
and slightly different inequalities for $p=2$ and $p \mid m$.


We have:
$$v\_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor v\_p(m^o-1) \ge v\_p((\frac{n}{o})!) + \lfloor \frac{n}{o} \rfloor = \sum\_{i \ge 1} \lfloor \frac{n}{op^i} \rfloor + \lfloor \frac{n}{o} \rfloor$$
$$= \sum\_{i \ge 1} \lfloor \frac{pn}{op^i} \rfloor \ge \sum\_{i \ge 1} \lfloor \frac{n}{p^i} \rfloor = v\_p(n!), $$
where the last inequality follows from the fact that $o<p$. This proves $(\*)$.


To deal with the exceptional primes ($2$ and divisors of $m$), just note that $P\_n(m)$ is divisible by $m^{\binom{n}{2}} (m-1)^n$ and that $v\_p(n!)\le n,\binom{n}{2}$ for every $p$ and $n$.


