Let $g$ be a primitive root modulo $p^e$ for some $p$ prime, $e\geq 1$, show that gcd$(g,p)=1$

Question

So far I've got:

Suppose gcd$(p,g)\neq 1$, so $p\mid g$ and hence $p^e\mid g^e$ so $g^e\equiv 0 $ (mod $p^e$)

Also $g^{p^{e-1}(p-1)}\equiv 1$ (mod $p^e)$ because $g$ is a primitive root.

Not sure where to go from here!

Answer 1

If $p\mid g$ then any power of $g$ is divisible by $p$, so you could never obtain $1\bmod {p^e}$.

Or: From $g^m\equiv 1\pmod {p^e}$ with $m\ge1$ we see $g^{m-1}\cdot g+kp^{e-1}\cdot p=1$ for some integer $k$, so $\gcd(g,p)\mid 1$.