Given:
- $p$ is a prime number.
- $r > 1$ is an integer.
- $m$ is an integer such that $p \nmid m$ (i.e., $p$ does not divide $m$).
- $x^m \equiv 1 \mod p^r$ and $x \equiv 1 \mod p$.
Goal: Prove that $x \equiv 1 \mod p^r$.
So far I am considering the binomial expansion of $x^m$, but am unsure where to go from here. I seem to keep going round in circles
