For an odd prime $p$, show that:
(a) Any primitive root of $p^2$ is also a primitive root of $p$
(b) Any primitive root of $p^n$ is also a primitive root of $p$
For part (a):
$r$ is a primitive root $\pmod {p^2}$ Suppose $r$ is not a primitive root $\pmod p$. Then there is some $n$ with $n|p−1$ by lagranges theorem. But then, $r^{np} \equiv 1 \pmod {p^2}$. Contradiction. Is this okay? I Didn't use the p being odd assumption here!
I have the general idea for (b):
Let $r$ be a primitive root of $p^n$ then: for $(a,p)=1$: $r^k\equiv a \pmod{p^n}$ $\Rightarrow r^k\equiv a \pmod{p}$ But how do we show the existence of $r$ & complete the proof?