Prove that the binomial coefficient is congruent to 0 mod p. Let $p$ be a prime number, and let $k$ be an integer such that $0<k<p$. Prove that the binomial coefficient ${p\choose k}\equiv 0\pmod p$.
How would I prove this?

Hints:
$$\binom pk=\frac{p!}{k!(p-k)!}=\frac{(p-k+1)(p-k+2)\cdot\ldots\cdot \overbrace{(p-1)}^{=p-k+(k-1)}}{1\cdot2\cdot\ldots\cdot k}\cdot p$$
But the above is an integer, and $\;k<p\;$ ...
