Suppose $p$ is a prime number not equal to $2$ and $m$ is a positive integer, where $$p \nmid 2^m - 1.$$ Prove that $$\sum_{n=1}^{p} n^m \equiv 0 \pmod{p}.$$
I first wrote out the sum, which gave me $$1^m + 2^m + \ldots + p^m \equiv 0 \pmod{p}.$$ However, I had no idea how to proceed from here.
$\sum_{n=1}^{p-1} n^m = 0\bmod p$ iff $p-1 \nmid m$.
If $p-1 \ | \ m$ then the sum is $p-1$
If $p-1 \nmid m$ then take $g$ such that $g^m \ne 0,1\bmod p$, we get that $$g^m\sum_{n=1}^{p-1} n^m = \sum_{n\ne 0\bmod p}(gn)^m = \sum_{n\ne 0\bmod p}n^m$$ which implies that the RHS is $0$.
