Bertrand-Chebyshev Theorem/Lemma 3

Lemma
Let $p$ be a prime number.

Let $p^k \divides \dbinom {2 n} n$.

Then $p^k \le 2 n$.

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

By De Polignac's Formula, the largest power of $p$ dividing $n!$ is $\ds \sum_{i \mathop \ge 1} \floor {\frac n {p^i} }$.

So: