Proving $\prod_{p\in \pi(m+1,2m)}p \leq {2m \choose m} $

Let $ \pi (m,n)$ denote the set of prime numbers in the interval $ [m,n]$

Show that $ \prod_{p\in \pi(m+1,2m)}p \leq {2m \choose m} $ .


My attempt: $ $ {2m \choose m} =\frac{m!(\prod_{p\in \pi(m+1,2m)}p)(\prod_{q\in [m+1,2m]-\pi(m+1,2m)}q)}{(m!)^2}=\frac{(\prod_{p\in \pi(m+1,2m)}p)(\prod_{q\in [m+1,2m]-\pi(m+1,2m)}q)}{m!}$ $

All I have to do now is explain why: $ $ \frac{\prod_{q\in [m+1,2m]-\pi(m+1,2m)}q}{m!} \geq 1$ $

but I’m struggling with a formal proof.