binomial coefficients - Proving that $nmid(nCr)$ for all $r$ ($1 leq r leq n-1$), only if $n$ is prime

I'm trying to prove that $n\mid(nCr)$ for all $r$ ($1 \leq r \leq n-1$) if and only if $n$ is prime.

Now proving that if $n$ is prime then $n\mid(nCr)$ is pretty easy, but how would you go about proving that $n\mid(nCr)$ only if $n$ is prime?

Could you show that if $n$ is not prime, then there exists an $r$ such that $n$ does not divide $nCr$? If so how would you go about doing that? I thought it would be easy but then I realized that for some $r, n\mid(nCr)$ even if $n$ isn't prime which made things a little bit more complicated.

I've managed to show that $n\mid(n-1)!$ for non-primes greater than or equal to $6$, which means that despite $n$ always canceling out with a product of factors in $r!(n-r)!$ in some cases, there will be another pair of factors that cancel out with the $n$ in $(n-1)!$, and sometimes there won't be. Maybe there's some kind of pattern but I can't find it unfortunately.

(I'd prefer a method that doesn't use undergraduate and above level maths)

Answer

Suppose n is not prime, and let $n=mp$ where $p$ is a prime. Then $n\not|\binom{n}{p}$, since
$$\binom{n}{p}=n \bigg[\frac{(n-1)!}{p!(n-p)!}\bigg]=n\bigg[\frac{(mp-1)!}{p!((m-1)p)!}\bigg]=n\bigg[\frac{(mp-1)(mp-2)(mp-3)\cdots((m-1)p+1)}{p!}\bigg],$$ where the expression in brackets is not an integer since
$(mp-1)(mp-2)(mp-3)\cdots((m-1)p+1)$ is not divisible by $p$, while $p!$ is.