A Couple Problems with Primes

1) Show that $\displaystyle 1^p + 2^p + 3^p + .... + (p-1)^p = 0$ (mod p) when p is an odd prime

2) Show that if *p* is a prime and *p* > 3, then $\displaystyle 2^{p-2} + 3^{p-2} + 6^{p-2} = 1$ (mod p)

For 1), I really don't understand which theorem or identity I'm supposed to put into use to prove this. Obviously it works every time I just don't understand how this works in terms of technicality.

2) is the same story. I don't see how it works and therefore I can't prove it formally.