How can one show that $\prod_{n<p\leq2n}p\leq C(2n,n)$?

Question

I am trying to rove that $\prod_{n<p\leq2n}p \leq C(2n,n) \leq 2^{2n}$, where $C(2n,n)= \frac{2n!}{n! n!}$ and $p$ is prime.

I can prove the second part by induction, but first part induction doesn't work.

I can see that $\prod_{n<p\leq 2n}p \leq \frac{2n!}{n!}$ but there is an extra $n!$ in the denominator.

Any suggestion?

Answer 1

Note that $p \mid (2n)!$ and since $p$ is prime and greater than $n$ it does not divide $n!$. From there you can derive that each $p$, and thus also the product of all, in fact divides the binomial coefficient.