n divides φ(aⁿ − 1)

Setup. Put $N=a^n-1\ge1$. If $N=1$ then $\varphi(1)=1$ and $1\mid1$, fine. Assume $N\ge2$.

Coprimality. $a\cdot a^{n-1}=a^n\equiv1\pmod N$, so $a$ is invertible modulo $N$.

The order is $n$. $a^n\equiv1\pmod N$. If $1\le k<n$, then $0<a^k-1<a^n-1=N$, so $N\nmid a^k-1$, i.e. $a^k\not\equiv1\pmod N$. Hence the order of $a$ modulo $N$ is exactly $n$.

Euler. $a^{\varphi(N)}\equiv1\pmod N$, so the order $n$ divides $\varphi(N)=\varphi(a^n-1)$.

Check. $a=2$, $n=4$: $N=15$, $\varphi(15)=8$, $4\mid8$ ✓. $a=3$, $n=3$: $N=26$, $\varphi(26)=12$, $3\mid12$ ✓.

Lesson. To show $n\mid\varphi(N)$, exhibit an element of order exactly $n$ modulo $N$.