In an answer here, it is said without proof that if $p$ is a prime and $n$ an integer, the order of $p$ in $(\Bbb Z/(p^n-1)\Bbb Z)^\times$ is $n$.
I tried to prove that, and it boils down to the equivalence $$p^n-1|p^k-1 \iff n|k$$.
While the $\Leftarrow$ part is not a problem, how can I prove the converse ?
You should look at it in the opposite way.
It is trivial to show that the order of $p$ in $(\mathbb Z/(p^n-1)\mathbb Z)^*$ is $n$. Because for $1 \leq d < n$, we have $1 < p^d < p^n-1$, in particular $p^d \neq 1 \mod{p^n-1}$.
And this shows the (a priori non-trivial) part of the equivalence $$p^n-1|p^k-1 \iff n|k$$
