Here is an olympiad-level problem on elementary number theory:
Let $a$ be an integer and $n$ a positive integer. Prove that \begin{align} \left(a-1\right)^n \cdot n! \mid a^{n-1} \prod_{i=1}^n \left(a^i-1\right) . \label{eq.darij1.1} \tag{1} \end{align}
There is a rather straightforward, if somewhat technical, way to prove this by comparing $p$-valuations (and using things like Euler's theorem, the lifting-the-exponent lemma, de Polignac's formula, and properties of the floor function). These are not what I am interested in right now.
Instead, I'm interested in what comes out if we cancel $\left(a-1\right)^n$ from both sides of \eqref{eq.darij1.1}. This transforms \eqref{eq.darij1.1} into the following equivalent form:
\begin{align} n! \mid a^{n-1} \prod_{i=1}^n \left(1+a+a^2+\cdots+a^{i-1}\right) . \label{eq.darij1.2} \tag{2} \end{align} The right hand side of this looks positively combinatorial: After all, if $S_n$ denotes the $n$-th symmetric group, and $\ell\left(w\right)$ denotes the Coxeter length (i.e., number of inversions) of any permutation $w \in S_n$, then a well-known formula (e.g., Proposition 5.3.5 in my An Introduction to Algebraic Combinatorics, as of 22 August 2021) says that \begin{align} \sum_{w \in S_n} x^{\ell\left(w\right)} = \prod_{i=1}^n \left(1+x+x^2+\cdots+x^{i-1}\right) . \end{align} This suggests interpreting the right hand side of \eqref{eq.darij1.2}, at least for nonnegative $a$ (but the case of negative $a$ can easily be reduced to the case of nonnegative $a$ by replacing $a$ by its remainder modulo $n!$), as a number of ways to (e.g.) pick a permutation $w \in S_n$, color each of its inversions by one of $a$ colors, and color each of the first $n-1$ positive integers by one of $a$ colors. If we could find a free action of the group $S_n$ on the set of such ways, then \eqref{eq.darij1.2} would follow.
More realistically, the set would have to be bijected to something more symmetric, on which $S_n$ acts more reasonably.
Question. Is there such a combinatorial proof of \eqref{eq.darij1.2}?
Note that something close to \eqref{eq.darij1.1} can be proved combinatorially when $a$ is a prime power, because in this case $a^{n\left(n-1\right)/2} \prod_{i=1}^n \left(a^i - 1\right)$ is the order of the group $\operatorname{GL}_n\left(\mathbb{F}_a\right)$ whereas $\left(a-1\right)^n \cdot n!$ is the size of its subgroup of monomial matrices (a wreath product of $\mathbb{F}_a^\times$ and $S_n$). This gives a weaker version of \eqref{eq.darij1.1} with a higher exponent on the $a$. However, this also only works when $a$ is a prime power, and as such arguments seem to have a bad track record of generalizing to general $a$, I'm inclined to consider this a dead-end.
