TITLE: Prove that the binomial coefficient is congruent to 0 mod p.
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: We have
$$(p-k)!k!\binom{p}{k} = p!=p(p-1)!$$
but $\gcd\left(p,(p-k)!k!\right)=1$ so by the Euclid's lemma $p|\binom{p}{k}$.

REPLY [2 votes]: There are $p$ chairs arranged uniformly around a circular table. We want to choose  $k$ of them. 
We say that two such choices $A$ and $B$ are equivalent if $B$ is obtainable from $A$ by a rotation. Since $p$ is prime, if $k$ is different from $0$ or $p$, there are precisely $p$ choices that are equivalent to $A$. 
Thus the set of choices of $k$ elements can be divided into families (equivalence classes) each of size $p$. It follows that the number of choices is divisible by $p$.