TITLE: Prove that , any primitive root $r$ of $p^n$ is also a primitive root of $p$
QUESTION [15 upvotes]: For an odd prime $p$, prove that any primitive root $r$ of $p^n$ is also a primitive root of  $p$

So I have assumed $r$ have order $k$ modulo $p$ , So $k|p-1$.Then if I am able to show that $p-1|k$ then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed.

REPLY [2 votes]: Exercise: If $X$ is a generating set for a group $G$ and $\phi:G\to H$ a homomorphism, then $\phi(X)$ is a generating set for the image of $G$ under $\phi$; in particular this means that if $\phi$ is surjective/onto then $\phi(X)$ is a generating set for $H$.
Now consider $G=(\Bbb Z/p^n\Bbb Z)^\times$ and $H=(\Bbb Z/p\Bbb Z)^\times$ with $\phi$ the mod $p$ map (see that it's well-defined and onto!) with a singleton generating set $X=\{a\}$ for $G$.