We already know that:
Let $X$ a finite set of size $n>0$ and $p$ a prime number. Let $\sigma$ be the circular permutation $(1,2,...,p)\in \mathfrak{S}_p$. If we consider the following group action: $\mathbb{Z}/p\mathbb{Z} \times X^p\longrightarrow X^p, \ (\bar{k},(a_1,...,a_p)) \mapsto (a_{\sigma^k(1)},...,a_{\sigma^k(p)})$ then by counting orbits under this action we can deduce that $p \mid (n^p-n)$.
But now I was wondering what would happen if we wanted $p^k \mid (n^{p^k}-n^{p^{k-1}})$ with $k>1$?
Maybe I could consider the same set $X$ of size $n>0$ and $\sigma$ the circular permuation $(1,2,...,p^k)\in \mathfrak{S}_{p^k}$. Then I could consider the following group action : $\mathbb{Z}/p^k\mathbb{Z} \times X^{p^k}\longrightarrow X^{p^k}, \ (\bar{k},(a_1,...,a_{p^k})) \mapsto (a_{\sigma^k(1)},...,a_{\sigma^k(p^k)})$.
If I reproduce the same method by counting orbits under that action, will I obtain the conclusion? Are there other alternatives which use a different group action?
Also, could we find the Euler's theorem with that method?
Thanks in advance!