Sum of hypercubes modulo $p$

Question

Let $p$ be a prime number and $n$ a natural number. What is $\sum_{i=1}^{p-1} i^n$ modulo $p$ ?

If $p>2$ and $n$ is odd it's easy to see that the sum is zero, but I don't see how to tackle this in general.

Answer 1

Let $g$ be a primitive root modulo $p$.
Then for every $1\leq i\leq p-1$ we have $i=g^{a_i}$ where $a_1,a_2,...,a_{p-1}$ is a rearrangement of the numbers $1,2,...,p-1$
So your sum is equal to $$g^{a_1\cdot n}+g^{a_2\cdot n}+...+g^{a_{p-1}\cdot n}=g^n+g^{2n}+...g^{(p-1)\cdot n}$$
which is equal to (using Fermat's little theorem) $$\frac{g^{np}-1}{g^n-1}-1\equiv\frac{g^n-1}{g^n-1}-1\equiv 0\pmod p$$