Let $k\ge2$ and p an odd prime number. If $g^{p-1}\equiv1[p]$ and $g^{p-1} \not\equiv 1 [p^2]$ then $g^{p^k-1} \equiv 1[p^k]$

Question

Let $k\ge2$ and p an odd prime number. If $$g^{p-1}\equiv1[p]$$ and
$$g^{p-1} \not\equiv 1 [p^2]$$ then $$g^{p^k-1} \equiv 1[p^k]$$ I want to use this lemma: $$g^n \equiv 1[p^k] \Rightarrow \phi(p^k)|n$$ Where $\phi$ is the euler's phi function. I tried to work with this lemma, but i don't get results.