Prove that $n$ devides $\phi(p^n-1)$

Question

Prove that $n$ devides $\phi(p^n-1)$ ($\phi(x)$ being the totient.)

I could not find anything about this particular question on the web, so I will share my argument here.

Answer 1

First, observe that $\phi(p^n-1)$ is the order of the group of automorphisms of a cyclic group of order $p^n-1$. I.e. $$\mid\Bbb{F}_{p^n}^*\mid=p^n-1$$ $$\mid Aut(\Bbb{F}_{p^n}^*)\mid=\phi(p^n-1)$$ $$Aut(\Bbb{F}_{p^n}^*)\ge Aut(\Bbb{F}_{p^n}/\Bbb{F}_p)=Gal(\Bbb{F}_{p^n}/\Bbb{F}_p)$$ $$\because Gal(\Bbb{F}_{p^n}/\Bbb{F}_p)\cong \Bbb{Z}/n\Bbb{Z}$$ $$\therefore n \mid \phi(p^n-1)$$