Show that $1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p$
Solution 1:
$$k^{p-1}\equiv 1 \text{ mod p} \text{ , } \forall p\nmid k$$ $$\sum_{k=1}^{p-1}k^{p-1}\equiv (p-1)\equiv -1 \text{ mod p}$$
Solution 2:
Hint: Apply Fermat's little theorem to each term, then sum.