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

Question

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.

Answer 1

Hint:

Any prime $p\in(m+1,2m)$ divides $(2m)!$ and does not divide $m!\cdot m!$. It follows that $p$ must divide $\binom{2m}{m}$.

Now consider the prime factorization of $\binom{2m}{m}$.