How can we show that
$\operatorname{ord}_{p}\left(\binom{2n}n\right) \le \frac{\log 2n}{\log p}$ where $p$ is a prime number and $n$ is a natural number
My attempt:
$2^{n} \le \prod _{k=1}^{n} \frac{k+n}{k} = \begin{pmatrix} 2n \\ n \end{pmatrix} = \frac{(2n)!}{n!n!} = \sum_{n\le 2n}p^{\operatorname{ord}_{p}(p)}$
$\Rightarrow \operatorname{ord}_{p}(p) \le \sum_{m=1}^{\infty}\left(\left\lfloor\frac{2n}{p^{m}}\right\rfloor-2\left\lfloor\frac{n}{p^{m}}\right\rfloor \right) \le \left\lfloor\frac{\log 2n}{\log p}\right\rfloor \le \frac{\log 2n}{\log p}$