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.

If r is a primitive root of prime p, we can show that either r is a primitive root of $p^2$ or (r+c.p) is a primitive root of $p^2$ where (c,p)=1.
We can also show that if g is a primitive root of both p and $p^2$, it is also a primitive root of $p^k$  where k is a natural number.
Clearly, there exists one r+c.p where p|c or  (c,p)=1 which is a primitive root of both p and $p^2$ =>r+c.p is a primitive root of $p^k$  where k is a natural number.
Each of $\phi(p-1)$ primitive root of p, will have associated (p-1) primitive roots which are congruent(mod p), but in-congruent(mod $p^2$)   of $p^2$, p(p-1) primitive roots  of $p^3$ and $p^{k-2}(p-1)$ primitive roots  of $p^k$ where k≥3.
Each primitive root of $p^k$ will be associated to p primitive roots of $p^{k+1}$ where k > 1.
