Contest problem on the divisibility of an integer by $n!$ This question is the $4$th question in part A of the Simon Marais Math Competition, $2022$ (held about $2$ weeks back).

Let $n$ be a positive integer, and $q \geq 3$ an odd integer such that every prime factor of $q$ is larger than $n$. Then $$
\frac{1}{n!} \prod_{i=1}^n \frac{q^i-1}{q-1}
$$
is an integer that has no prime factor in common with $\frac{q-1}{2}$.

This is a competition for undergraduate students, therefore I expected a typical olympiad technique to be used here. That $q-1$ divides $q^i-1$ is obvious. We want to show that $n!$ divides this quotient.
Let $p \leq n$ be odd, and $p | q-1$. By Lifting the Exponent, $\nu_p(\frac{q^i-1}{q-1}) = \nu_p(q^i-1)-\nu_p(q-1) = \nu_p(i)$. Thus, $\nu_p(\prod_{i=1}^n \frac{q^i-1}{q-1}) = \sum_{i=1}^n \nu_p(i) = \nu_p(n!)$. If $p=2$ then $p$ divides $q-1$, and LTE for $p=2$ does the trick again.
However, if $p \nmid q-1$ then as $p \nmid q$, $q^{k(p-1)} \equiv 1 \mod p$ for all $k$... and I'm not sure what to do. It is not clear to me that $\nu_p(\prod_{i=1}^n(q^i-1)) \geq \nu_p(n!)$. Certainly it is true that the number of $i$ such that $q^i-1$ is a multiple of $p$ is bigger than the number of multiples of $p$ less than $n$. However, accounting for $p^2,p^3$ etc. is the problem. I'd want to know if someone can continue and finish this approach.
I assume that it's enough to show that it is an integer. Once that occurs, it should not be difficult to show that $\nu_{p}((q-1)/2)>0$ implies that $p$ doesn't divide that fraction, using LTE and breaking into cases $p=2,p \neq 2$.
 A: Let $d = \text{ord}_p(q)$ be the smallest positive integer $d$ such that $q^d \equiv 1 \bmod p$, so that $q^i \equiv 1 \bmod p$ iff $d \mid i$. By Fermat's little theorem, $d \mid p-1$, so in particular $d \le p-1$. This gives, without even using LTE, that
$$\nu_p \left( \prod_{i=1}^n \frac{q^i - 1}{q - 1} \right) \ge \left\lfloor \frac{n}{d} \right\rfloor \ge \left\lfloor \frac{n}{p-1} \right\rfloor \ge \nu_p(n!).$$
I haven't thought about the second part of the problem. I can't resist mentioning that if $q$ is a prime power there is a much more direct proof of this divisibility: in this case it follows by Lagrange's theorem applied to the fact that the general linear group $GL_n(\mathbb{F}_q)$, which has order $q^{{n \choose 2}} \prod_{i=1}^n (q^i - 1)$, has a subgroup of order $(q - 1)^n n!$, namely the generalized permutation matrices or monomial matrices, which abstractly are the wreath product $\mathbb{F}_q^{\times} \wr S_n$. If every prime factor of $q$ is larger than $n$ then $(q - 1)^n n!$ is relatively prime to $q^{ {n \choose 2} }$ so that factor can be ignored.
I also can't resist mentioning some more interesting facts. The polynomial $[n]_q! = \prod_{i=1}^n \frac{q^i - 1}{q - 1}$ is known as the $q$-factorial and many fun things are known about it. Its coefficients have a combinatorial interpretation in terms of permutations, and also as the Betti numbers of a manifold called a flag variety. If $q$ is a prime power it counts the number of complete flags in the vector space $\mathbb{F}_q^n$, and the relationship between this count and the fact about Betti numbers comes from the Weil conjectures.
