Prove that if $p$ is a prime number then $\binom{p-1}{k}\equiv (-1)^k\pmod{p}$ 
Prove that if $p$ is a prime number then $\binom{p-1}{k}\equiv (-1)^k\pmod{p}$. What can be said about $\binom{p+1}{k} \pmod{p}$?

I thought about expanding $\dbinom{p-1}{k} = \dfrac{(p-1)!}{k!(p-1-k)!}$, but I don't see how that helps. Is there an easier way?

If $k=0$ then it is trivially true. Otherwise:
$$\frac{(p-1)!}{k!(p-1-k)!}\equiv\frac{(p-1)\ldots(p-k)}{k!}\equiv\frac{(-1)\ldots(-k)}{1\ldots k}\equiv (-1)^k \frac{k!}{k!}\equiv(-1)^k,$$
with the appearance of divisions making sense because $\mathbb{Z}_p$ is a field.
