Prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$. 
For an odd prime, prove that a primitive root of $p^2$ is also a primitive root of $p^n$ for $n>1$.  

I have proved the other way round that any primitive root of $p^n$ is also a primitive root of $p$ but I have not been able to solve this one. I have tried the usual things that is I have assumed the contrary that there does not exist the primitive root following the above condition and then proceeded but couldn't solve it.
Please help.

We know from here, if $ord_pa=d, ord_{(p^2)}a= d$ or $pd$
If $a$ is a primitive root $\pmod {p^2}, ord_{(p^2)}a=\phi(p^2)=p(p-1)$ 
Then $ord_pa$ can be $p-1$ or $p(p-1)$
But as $ord_pa<p, ord_pa$ must be $p-1=\phi(p)\implies a$ is a primitive root $\pmod p$
Again from here, $$ord_{(p^s)}(a)=d, ord_{p^{(s+1)}}(a)=pd,\text{ then } ord_{p^{(s+2)}}(a)=p^2d$$
So, as $ord_pa=p-1,ord_{(p^2)}a=p(p-1); ord_{(p^3)}a$ will be $p\cdot p(p-1)=\phi(p^3)$
Again as,  $ord_{(p^2)}a=p(p-1) ord_{(p^3)}a=p\cdot p(p-1);ord_{(p^4)}a$ will be $p\cdot p^2(p-1)=\phi(p^4)$
