Prove or disprove that $\phi(a^n - 1)$ is divisible by n I have a proof for the case of $a$ being prime I believe, I think this is also true for $a$ composite since I ran a test for the first $100$ numbers over the first $100$ values of $n$ and it seems to be true.
Anyway the proof for 'a' prime is indirect and through finite field theory. 
$ \phi(a^n - 1)$ is the number of primitive roots in $GF(a^n)$ and they have irreducible polynomials of degree n over $GF(a)$ and all the roots of that irreducible equation have to be primitive roots. So the irreducible equations partition the set of primitive roots into sets of $n$ elements which means $ \phi(a^n - 1)$ is divible by $n$.
Any help would be appreciated.

If $a > 1$ is an integer, then the order of $a$ in $\mathbb{Z}_{a^n-1}^*$ is $n$.
