Let $\phi(n)$ be the Euler phi-function. If $a>1$ is an integer, then what is the remainder when $\phi(a^n - 1)$ is divided by $n$ in accordance with the Euclidean algorithm?
The group $(\mathbb{Z}/(a^n-1)\mathbb{Z})^\times$ has an element of order $n$, namely $\overline{a}$, because obviously $$a^n\equiv 1\bmod (a^n-1),$$ and $a^k\not\equiv 1\bmod (a^n-1)$ for any $0<k<n$ because $1<a^k-1<a^n-1$ for any $0<k<n$.
Because the group has an element of order $n$, the order of the group is divisible by $n$.
Since $\varphi(N)=|(\mathbb{Z}/N\mathbb{Z})^\times|$, we have that $\varphi(a^n-1)$ is divisible by $n$. Thus the remainder is $0$.