Let $p$ be a prime, and let $N=\left(\begin{array}{c}2 n \\ n\end{array}\right)$ for some positive integer $n$.

Show that if a prime power $p^{k}$ divides $N$ for some $k \geqslant 1$, then $p^{k} \leqslant 2 n$.

Given a positive real $x$, define $\psi(x)=\sum_{n \leqslant x} \Lambda(n)$, where $\Lambda(n)$ is the von Mangoldt function, taking the value $\log p$ if $n=p^{k}$ for some prime $p$ and integer $k \geqslant 1$, and 0 otherwise. Show that

$$\psi(x)=\sum_{p \leqslant x, p \text { prime }}\left\lfloor\frac{\log x}{\log p}\right\rfloor \log p$$

Deduce that for all integers $n>1, \psi(2 n) \geqslant n \log 2$.