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

For any $a$ relatively prime to $p^n$, there is an integer $k$ such that $r^k\equiv a \pmod{p^n}$ and hence such that $r^k \equiv a\pmod{p}$. Thus $r$ is a primitive root of $p$.

Note that an integer $r$ with $\gcd(r,p)=1$ is a primitive root modulo $p^k$ when the smallest $b$ such that $r^b\equiv1\bmod p^k$ is $b=p^{k-1}(p-1)$.

Suppose that $r$ is not a primitive root modulo $p$, so there is some $b<p-1$ such that $r^b\equiv 1\bmod p$.

In other words, there is some integer $t$ such that $r^b=1+pt$.

Then of course we have that $p^{n-1}b<p^{n-1}(p-1)$, and $$r^{p^{n-1}b}\equiv 1\bmod p^n$$ because of

the binomial theorem.

(mouse over to reveal spoilers)

$\begin{eqnarray}{\bf Hint}\ \ \ \ \rm a\ \ is\, \ coprime\, \ to\ \ {\bf\color{#C00}p}\,\ &\Longrightarrow&\rm\ \ \ a\ \ is\ \ coprime\ \ to\ \ {\bf\color{#90f}{p^n}}\\ & & \qquad\qquad{ \Downarrow}\\ \smash{\rm r^k\equiv a\ \ (mod\,\ {\bf\color{#C00}p})}\, &\smash{\overset{\ \ \rm\color{#0a0}{CP}}\Longleftarrow}&\rm \smash{\exists\,k\!:\ r^k\equiv a\ \ (mod\,\ {\bf\color{#90f}{p^n}})} \end{eqnarray}$

where we employed the result $\,\rm \color{#0a0}{CP} = $ Congruences Persist mod factors of the modulus.