TITLE: congruent to mod p $1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$
QUESTION [5 upvotes]: Let $p$ be an odd prime.How to prove that
$$1^{p-2}+2^{p-2}+\cdots+\left(\frac{p-1}{2}\right)^{p-2}\equiv\frac{2-2^p}{p}\pmod p.$$

REPLY [2 votes]: $$2^p=(1+1)^p=\sum_{k=0}^p\binom {p}{k}=\left[\binom{p}{0}+\binom{p}{p}\right]+\ldots+\left[\binom{p}{\frac{p-1}{2}}+\binom{p}{\frac{p+1}{2}}\right]=$$
$$=2\left[\binom{p}{0}+\ldots+\binom{p}{\frac{p-1}{2}}\right]$$
and etc.