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.

By a combinatorial argument, or by manipulating factorials, we have
$$i\binom{p}{i}=p\binom{p-1}{i-1}\ .$$
Since $\binom{p-1}{i-1}$ is an integer,
$$p\mid i\binom{p}{i}\ .$$
But $p$ is prime and $1\le i<p$, so $p$ and $i$ have no common factor, so
$$p\mid \binom{p}{i}\ .$$
For the case when $p$ is not prime, just take $p=4$ and try out various values of $i$.
