How to show $\binom{2n}{n} \ge \prod_{n < p \le 2n} p $?

Question

What is the best way to show \begin{equation} \binom{2n}{n} \ge \prod_{n < p \le 2n} p \end{equation} for prime $p$.

I know that $ 2^{2n} = (1+1)^{2n} \ge \binom{2n}{n}$.

and \begin{equation} \binom{2n}{n} = \frac{2n(2n-1)...(n+1)}{n!} \end{equation}

Answer 1

Actually something stronger holds. It turns out that the product in question divides the binomial coefficient.

Why is this? Well we know that the binomial coefficient is a positive integer. Also any prime $n<p\leq 2n$ must divide this positive integer since $p|(2n)!$ and $p\nmid n!$.