Prove that , any primitive root $r$ of $p^n$ is also a primitive root of $p$ 
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.

$\begin{eqnarray}{\bf Hint}\ \ \ \   \rm a\ \ is\,  \ coprime\, \ to\ \ {\bf\color{#C00}p}\,\  &\Longrightarrow&\rm\ \ \ a\ \ is\  \ coprime\ \ to\ \ {\bf\color{#90f}{p^n}}\\ 
& & \qquad\qquad{ \Downarrow}\\  
\smash{\rm r^k\equiv a\ \ (mod\,\ {\bf\color{#C00}p})}\, &\smash{\overset{\ \ \rm\color{#0a0}{CP}}\Longleftarrow}&\rm  \smash{\exists\,k\!:\ r^k\equiv a\ \ (mod\,\ {\bf\color{#90f}{p^n}})}
\end{eqnarray}$
where we employed the result  $\,\rm \color{#0a0}{CP} = $ Congruences Persist mod factors of the modulus.
