What do you think?

Can you prove that if $p$ is prime and $n$ is a natural number: $${n \choose p} \equiv \left[ \frac{n}{p} \right](\text{mod } p)$$

Note that $\left[ \frac{n}{p} \right]$ denotes the floor of $\frac{n}{p}$ which is the integer part of it.

Also note that $a \equiv b (\text{mod }n)$ means that $(a-b)$ is a multiple of $n$