If $p$ is a prime satisfying $n<p<2n$, show that $\binom{2n}{n}\equiv 0\mod p$.
See that $$\binom{2n}{n}=\frac{(2n)!}{n!n!}=\frac{2n\cdot (2n-1)\cdot (2n-2)\cdots (n+1)}{n\cdots 3\cdot 2\cdot 1}$$
$p$ lies between $n$ and $2n$ so $p$ divides $2n\cdot (2n-1)\cdot (2n-2)\cdots (n+1)$.
Now, denominator has $2n$ so, it cancels with $2n$ in numerator leaving
\begin{align} \frac{(2n-1)\cdot (2n-2)\cdots (n+1)}{(n-1)\cdots 3\cdot 1} &= \frac{(2n-1)\cdot 2(n-1)\cdots (n+1)}{(n-1)\cdots 3\cdot 1} \\[0.3cm] &= \frac{(2n-1)\cdot 2\cdot (2n-3)\cdot 2\cdots (n+1)}{(n-2)\cdots 3\cdot 1} \end{align}
I am not very sure how to proceed..
I see that $n-2$ in the denominator gets cancelled with $2n-4$ in the numerator and so on... As $p$ is already out of business, there is no $p$ and we are cancelling $2(n-k)$ so they are not primes...
It might be easiest to use the fact from number theory that $p\mid ab$ implies $p\mid a$ or $p\mid b$. In this case let $a={2n\choose n}$ and $b=n!n!$. Then $ab=(2n)!$ which is divisible by any $p$ less than $2n$, while $b$ is not divisible by any $p$ greater than $n$.