$ 1 + 2^{p-2} + 3^{p-2} + \cdots + (p-1)^{p-2} \equiv 0\pmod p $ with $ p $ odd prime Let $ m $, $ m+1 $, $ m+2 $, $ \dots $, $ m+p-1 $ be an integers and let $ p $ be an odd prime. I want to show that
$$
m + (m+1)^{p-2} + (m+2)^{p-2} + \cdots + (m+p-1)^{p-2} \equiv 0\pmod p.
$$
This comes down to showing that
$$
1 + 2^{p-2} + 3^{p-2} + \cdots + (p-1)^{p-2} \equiv 0\pmod p,
$$
because by the pigeonhole principle, without loss of generality we may assume that $ m \equiv 0 \pmod p$, and then $ m+i \equiv i \pmod p$ for $ i = 1,2,\dots,p-1 $
 A: Hint:
Work in the field $\,\mathbf F_p=\mathbf Z/p\mathbf Z$ and observe that by Fermat, for any $k\in[1..\,p-1]$, one has 
$$k^{p-2}=k^{-1}.$$
Now, in this field, the map
\begin{align}
\mathbf F_p^{\times}&\longrightarrow \mathbf F_p^{\times} \\
k&\longmapsto k^{-1}
\end{align}
is an automorphism of the (multiplicative) group of non-zero elements. 
Therefore the given sum is just, modulo $p$, the sum $\;1+2+\dots+(p-1)$ in another order. Can you finish the proof?
A: Here's a nice alternative to the approach suggested in the comments by user rtybase:
If you are familiar with a bit of group theory, you may know that for every odd prime $p$, the group $(\Bbb{Z}/p\Bbb{Z})^{\times}$ of units  modulo $p$ is cyclic of order $p-1$, written multiplicatively. So taking $k$-th powers yields an automorphism of the group whenever $k$ is coprime to $p-1$. It follows that
$$1^k+2^k+\ldots+(p-1)^k\equiv1+2+\ldots+(p-1)\pmod{p}.$$
A: Yet another way of computing the sum, or (following @Servaes) even more general $$S_k=\sum_{a\in\mathbb{F}_q^{\times}}a^k$$ where $\mathbb{F}_q$ is a finite field with $q$ elements ($q$ is a power of a prime; this includes the case of $q=p$) and $k$ is any integer, is as follows. Assume (again) we know that $\mathbb{F}_q^{\times}$ is cyclic, and let $g$ be a generator of this group. If $k$ is a multiple of $q-1$, then each term of $S_k$ is $1$, so $S_k=q-1=-1$ in this case. Otherwise $g^k\neq 1$, and since $a\mapsto ga$ is a bijection of the group onto itself, we have $$S_k=\sum_{a\in\mathbb{F}_q^{\times}}(ga)^k=g^k S_k,$$ hence $(g^k-1)S_k=0$ and $S_k=0$.
