I have this problem assigned for homework, and I know there is already a post of the same problem, but I wanted to see how to solve it with a different method:
Let $p$ be an odd prime. Prove that any primitive root $r$ of $p^n$ is also a primitive root of $p$.
The hint that is given is: Let $r$ have order $k$ modulo $p$. Show that $r^{pk}\equiv 1\pmod{p^2},\dots,r^{p^{n-1}k}\equiv 1\pmod{p^n}$ and hence $\phi(p^n)\mid p^{n-1}k$.
Proof
Let $o_p(r)=k$.
(Now I am unsure how to prove that this implies $r^{p^{n-1}k}\equiv 1\pmod{p^n}$.)
Now, since $r^{p^{n-1}k}\equiv 1\pmod{p^n}$, then we must have $o_{p^n}(r)\mid p^{n-1}k$. But $r$ is a primitive root of $p^n$, so we have $o_{p^n}(r)=\phi(p^n)\implies\phi(p^n)\mid p^{n-1}k\implies p^{n-1}(p-1)\mid p^{n-1}k\implies (p-1)\mid k$.
But $r^k\equiv 1\pmod{p}\implies k\mid\phi(p)\implies k\mid p-1$, so we must have $k=(p-1)$, which means $o_p(r)=p-1\implies r$ is a primitive root of $p$. $\blacksquare$
Thanks in advance!
To take a totally different approach, choose some nonzero $\ell \pmod p$. Since $r$ is a primitive root mod $p^n$, we know there is some $k$ so that $r^k \equiv \ell \pmod {p^n}$. But then $r^k \equiv \ell \pmod {p}$ as well.
So we see directly that each residue class mod $p$ can be written through powers of $r$, and $r$ is a primitive root mod $p$ as well.