$g$ a primitive root modulo $p^2$ for $p$ an odd prime $\Rightarrow$ $g$ a primitive root modulo $p^k$ for every $k \in \mathbb{Z}^+$

Question

I've been trying to prove this result to no avail. Previously I managed to prove that if $p$ is an odd prime and $g$ a primitive root modulo $p$, then $g$ or $g + p$ is a primitive root modulo $p^2$. I thought I could maybe prove this result by induction on $k \in \mathbb{Z}^+$ plus the observation that every $k$ is either even or odd, hence $= 2m$ or $= 2m + 1$ for some lesser $m$ (so $p^k$ is either $p^{2m}$ or $p \cdot p^{2m}$), but I haven't really figured out how I can use this property. Any help would be much thanked.