For any prime number $p$ and natural number $i < p$, prove that $p$ divides ${p \choose i}$. For any prime number $p$ and natural number $i < p$, prove that $p$ divides ${p \choose i}$.
Also, what happens when $p$ is not a prime. Is this still true?
I tried writing out the formula for combination but couldn't get further.

It is not true in general (look for a counterexample at $p=4$). For $p$ prime though, you have
$$
{p\choose i} = \frac{p!}{i!(p-i)!} \in \mathbb{N}
$$
So
$$
p \mid {p\choose i}i!(p-i)!
$$
However, $p\nmid i!, p\nmid (p-i)!$. So $p\mid {p\choose i}$.
