If $p$ is a prime satisfying $n<p<2n$ then $\binom{2n}{n}\equiv 0 \pmod p$.

Question

We have that $\binom{2n}{n}= (2n)!/(n!)^2=k$ for some natural number $k$ as binomial coefficients are integers and in this case $k$ must be positive. Then $(n!)^2 \mid (2n)!$ but the $\gcd((n!)^2,p)) = 1$. This means that $(n!)^2 \mid ((2n)!/p)$. So $(2n)!/p = k(n!)^2$ which implies that $pk = (2n)!/(n!)^2 = \binom{2n}{n}$ and the congruence follows.

Answer 1

That proof looks fine to me.

You write $$ (n!)^2 \mid ((2n)!/p) $$ but you probably need to first explain by the right hand side is an integer. (It clearly IS, because $p$ is one of the factors in $(2n)!$, but you should probably SAY so.)