Using Fermat's Little Theorem Prove if $p$ is prime, prove $1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}$ Using Fermat's Theorem prove if $p$ is prime, prove $1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}$
The two definitions of Fermat's Little Theorem is $a^p \equiv a \bmod{p}$ and $a^{p-1} \equiv 1 \bmod{p}$ but I don't know how to use this solve the problem 

Since $1^p \equiv 1$, and $2^p \equiv 2$, $\ldots$ $(p-1)^p \equiv p-1$, we have that
$$1^p + 2^p + \ldots (p-1)^p \equiv 1 + 2 + \ldots + p-1$$
However, we know this sum: $\sum_{i = 1}^{p-1} i = \frac{p (p-1)}{2}$, which for odd primes would be a multiple of $p$, and we are done by reducing modulo $p$, then we only must check the case for $2$, which doesn't hold, and it's awkward...
