Combinatorial Proof for a $ p\mid\binom{p}{k} \ \ \ \ \ 0I'm looking for a combinatorial proof to the following statement:
$$ p\mid\binom{p}{k} \ \ \ , \ \ 0<k<p \ \ \ \ \ \ \text{and} \ \  p \   \text{is prime}.$$ 
Thank you.

If you count the number of $k$-element subsets of a $p$-element set that contain a given element and then sum over all $p$ elements, you get $k\binom pk$, since each element is in $k$ of the $\binom pk$ subsets. This is $p$ times the count for a single element, and since $p\nmid k$ for $0\lt k\lt p$, the factor $p$ must be in $\binom pk$.
