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.

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.

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