Bertrand-Chebyshev Theorem/Lemma 3

Lemma
If $p$ is a prime number and $p^k \mathop \backslash \dbinom {2 n} n$, then $p^k \le 2 n$.

Proof
Let $l$ be the largest power of $p$ with $p^l \le 2 n$.

By Multiplicity of Prime Factor in Factorial, the largest power of $p$ dividing $n!$ is $\displaystyle \sum_{i \mathop \ge 1} \left \lfloor{\frac n {p^i}}\right \rfloor$.

So: