TITLE: Prove $\binom{p-1}{k} \equiv (-1)^k\pmod p$
QUESTION [9 upvotes]: Prove that if $p$ is an odd prime and $k$ is an integer satisfying  $1\leq k \leq p-1$,then the binomial coefficient
   $$\binom{p-1}{k} \equiv (-1)^k\pmod p$$

I have tried basic things like expanding the left hand side to $\frac{(p-1)(p-2).........(p-k)}{k!}$ but couldn't get far enough.

REPLY [4 votes]: Recall that we have $(x + 1)^p \equiv x^p + 1 \bmod p$. The ring $\mathbb{F}_p[x]$ is an integral domain, so we can divide and it follows that
$$(x + 1)^{p-1} \equiv \frac{x^p + 1}{x + 1} \equiv x^{p-1} - x^{p-2} \pm ... - x + 1 \bmod p.$$