Isn't this almost trivial by LTE? Take a prime $p \leq n$. Let $d = \text{ord}_p(a) \le p-1$. Then: $v_p \left ( \prod_{k=1}^{n-1} (a^k - 1) \right ) \ge \sum_{k=0}^\infty \left \lfloor \frac{n-1}{p^k(p-1)} \right \rfloor$, which is at least as large as $v_p(n!) = \sum_{k=1}^\infty \left \lfloor \frac{n}{p^k} \right \rfloor$ because $\frac{n-1}{p-1} \geq \frac{n}{p}$, thus $\left \lfloor \frac{n-1}{p^{k-1}(p-1)} \right \rfloor \ge \left \lfloor \frac{n}{p^k} \right \rfloor$ and then summing through $k$ we get the result, as $v_p \left ( \prod_{k=1}^{n-1} (a^k - 1) \right ) \ge v_p(n!)$ for every $p\mid n!$. [mod: I have edited the post, to also include the possible case $p=n$.]

It is almost trivial by any account. Take a prime $2\leq p \leq n$. Then $v_p(n!) = \sum_{k=1}^\infty \left \lfloor \frac{n}{p^k} \right \rfloor < \sum_{k=1}^\infty \frac{n}{p^k} = \frac {n} {p-1}$. It is immediate that this implies $v_p(n!) \leq \left \lfloor \frac{n-1}{p-1} \right \rfloor$. On the other hand, since $\gcd(p,a)=1$, it follows from Fermat's Little theorem that at least $\left \lfloor \frac{n-1}{p-1} \right \rfloor$ factors of the form $a^k-1$, $1\leq k \leq n-1$, are divisible by $p$.