Prove that $q^i \equiv 1 \pmod {n!}$ for all $q, n \in \mathbb{Z^+}$ where the prime factors of $q$ are greater than $n$ The question originally states:

Let $n$ and $q$ are positive integers, such that all prime divisors of $q$ are greater than $n$. Show that $$(q-1)(q^2-1)(q^3-1)...(q^{n-1}-1) \equiv 0 \pmod {n!}$$

I decided it would be easier to attempt and prove that there will always be at least one $q^i \equiv 1 \pmod {n!}$ but failed. I was hoping to get some assistance (maybe a hint) on this question since I never dealt with factorial and a modular question.
 A: For each prime natural number $p\leq n$, the largest nonnegative integer $r$ such that $p^r$ divides $n!$ is $r=e_p(n)$, where
$$e_p(n):=\sum_{k=1}^\infty\,\left\lfloor\frac{n}{p^k}\right\rfloor\,.$$
We shall prove that if $E_{p,q}(n)$ is the largest nonnegative integer $r$ such that $p^r$ divides $\prod\limits_{k=1}^{n-1}\,\left(q^k-1\right)$, then
$$e_p(n)\leq E_{p,q}(n)\,.$$
Let $v_p$ be the $p$-adic valuation (that is, $e_p(n)=v_p(n!)$ and $E_{p,q}(n)=v_p\left(\prod\limits_{k=1}^{n-1}\,\left(q^{k}-1\right)\right)$).  We have for every positive integer $k$ that
$$v_p\left(q^{k(p-1)}-1\right)=v_p\left(q^{p-1}-1\right)+v_p(k)\geq 1+v_p(k)$$
by the Lifting-the-Exponent Lemma.  (The inequality $v_p\left(q^{p-1}-1\right)\geq 1$ is due to Fermat's Little Theorem.)
In particular, we see that
$$E_{p,q}(n)\geq \sum_{k=1}^{\left\lfloor\frac{n-1}{p-1}\right\rfloor}\,v_p\left(q^{k(p-1)}-1\right)\geq \sum_{k=1}^{\left\lfloor\frac{n-1}{p-1}\right\rfloor}\,\left(1+v_p(k)\right)\,.$$
Since $e_p(n)=\displaystyle \sum_{k=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\,v_p(pk)= \sum_{k=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\,\left(1+v_p(k)\right)$ and $\left\lfloor\dfrac{n-1}{p-1}\right\rfloor \geq \left\lfloor\dfrac{n}{p}\right\rfloor$, we conclude that
$$E_{p,q}(n)\geq \sum_{k=1}^{\left\lfloor\frac{n}{p}\right\rfloor}\,\left(1+v_p(k)\right)=e_p(n)\,.$$

It is however not true that $q^i\equiv 1\pmod{n!}$ for some $i\in\{1,2,\ldots,n\}$.  For example, take $n:=6$ and $q:=11$.  Then,
$$q^1-1=10\not\equiv0\pmod{6!}\,,$$
$$q^2-1=120\not\equiv0\pmod{6!}\,,$$
$$q^3-1=1330\equiv 610\equiv\not\equiv0\pmod{6!}\,,$$
$$q^4-1=14640\equiv 240\not\equiv0\pmod{6!}\,,$$
$$q^5-1=161050\equiv 490 \not\equiv0\pmod{6!}\,.$$
and
$$q^6-1=1 1771560\equiv 360\not\equiv 0\pmod{6!}\,.$$
