I know that for an odd prime $p$ and $a,b \in \mathbb{Z}$ : $a \equiv b \mod p^k \Rightarrow a^p \equiv b^p \mod p^{k+1}$.
I am trying to prove the converse, mainly that for $k>1$ and an odd prime $p$ we have : $a^p \equiv 1 \mod p^k \Rightarrow a \equiv 1 \mod p^{k-1}$.
Any help?
You have $p^k\mid a^p-1$. Now, suppose first $k=2$. Note also that, $a^p-1\equiv a-1\pmod{p}$, by Fermat, and thus, $p\mid a-1$. Thus, the statement is valid for $k=2$. Next, let $k\geqslant 3$. We have $p^k \mid a^p-1$. Let $p^2\mid a^p-1$. Recall also that $a^p-1=(a-1)(a^{p-1}+\cdots+1)$. Now, observe that, $a^{p-1}+\cdots+1\equiv p\pmod{p^2}$. Thus, the largest power of $p$ dividing $a^{p-1}+\cdots+1$ is one, hence, if $p^k\mid a^p-1$, we necessarily have $p^{k-1}\mid a-1$.
Now, suppose $p\mid \mid a-1$, that is, $p\mid a-1$ and $p^2\nmid a-1$. I now claim that $p^k\mid a^p-1$ is impossible for $k\geqslant 3$. To see this, let $a=p\ell+1$ with $p\nmid \ell$. Using binomial expansion: $$ a^p-1=(p\ell+1)^p -1=\sum_{k=2}^p \binom{p}{k}(p\ell)^k + p^2\ell. $$ Since $p\mid \binom{p}{k}$ for every $1\leqslant k\leqslant p-1$, we have $\sum_{k=2}^p \binom{p}{k}(p\ell)^k\equiv 0\pmod{p^3}$. Hence, the overall object is $a^p-1\equiv p^2\ell\pmod{p^3}$, and since $p\nmid \ell$, this forces $k=2$, which is the case we have handled in the beginning.
