While preparing for an exam, I am currently working on the following task:
Let $n$ be a natural number, and $\zeta_n$ a primitive $n$th root of unity. Show that \begin{equation*} \sum_{j=1}^{n} \binom{\zeta_n^j}{n} = \frac{1}{(n-1)!} \end{equation*}
It can be assumed to be known that $\sum_{k=1}^{n} \zeta_n^k = 1$ for $n=1$ and $\sum_{k=1}^{n} \zeta_n^k = 0$ otherwise.
I would be grateful for some simple hints for proving this equation. The task was given in a couse on combinatorics, and its solution is obviously somehow related to the Stirling numbers of the first kind. Please do not post the full solution.