Every prime power $p^k$ that divides $\binom{2m}{m}$ is smaller than or equal to $2m$

Hint: Let $\alpha_{n, p}$ be the largest natural number such that $p^{\alpha_{n, p}}\mid n!$ for $n\in\mathbb{N}$ and $p$ prime. Then $$\alpha_{n, p}=\sum_{k=1}^{\infty}{\left[\frac{n}{p^k}\right]},$$ where $[x]$ is the integer part of $x$.