Problem - 4232
Let $p$ be a prime and $k$ be a positive integer less than $p$. Show that $\binom{p}{k} \equiv 0 \pmod{p}$.
Because $p$ is a prime and $k < p$, we must have $p\not\mid k!$. Meanwhile, because $k > 0$, we have $(p-k) < p$ which means $p\not\mid (p-k)!$.
However, $\binom{p}{k}= \frac{p!}{k!(p-k)!}$ is an integer. The denominator, $p!$, is obviously a multiple of $p$, but the numerator is not. Hence, the result must be a multiple of $p$.