Problem - 4229
Let $p$ is an odd prime, compute $1^{p}+2^{p}+3^{p}+\cdots+(p-1)^{p}\pmod{p}$.
By Fermat's little theorem, we have $$\begin{array}{ll} \therefore\quad &1^{p}+2^{p}+3^{p}+\cdots+(p-1)^{p}\\ \equiv& 1+2+3+\cdots + (p-1)\\ \equiv& p(p-1)/2\\ \equiv& \boxed{0}\pmod{p} \end{array}$$