Let $p$ be a prime and $n$ a positive integer. Suppose there is a unique irreducible polynomial of degree $n$ over $\mathbb{F}_p$. Prove that $n = \phi(p^n-1)$, where $\phi(k)$ is Euler's function.
This one is particularly tough for me. I'm not sure how to approach this. Would it help that $\phi(p^n-1)$ is equal to $|\mathbb{Z_{p^n-1}}^{\times}|$, i.e., $\phi(p^n-1)$ counts the number of units in $\mathbb{Z}_{p^n-1}$ ? I thought of possibly showing that the unique irreducible polynomial described has the desired number of roots. But this isn't guaranteed in $\mathbb{F}_p$ alone, but rather, possibly in some extension field of $\mathbb{F}_p$, so I'm not sure how to go about this.
Any help would be appreciated.
Thanks!
In general, if $\alpha$ is a generator of $\mathbb{F}_{p^n}^\times$, then the Frobenius map $a \mapsto a^p$ permutes the roots of its minimal polynomial $m$, so $\alpha, \alpha^p, \ldots, \alpha^{p^{n-1}}$ are roots of $m$, which are distinct (why?). But these are all generators of $\mathbb{F}_{p^n}^\times$, so $n \le \phi(p^n-1)$, since $\mathbb{F}_{p^n}^\times$ has $\phi(p^n-1)$ generators.
Now if $\mathbb{F}_p$ has a unique minimal polynomial $f$ of degree $n$, then the minimal polynomial of each generator of $\mathbb{F}_{p^n}^\times$ must be $f$ (since it has degree $n$). But that implies that each generator is a root of $f$, so $\phi(p^n-1) \le n$.