Show that the exact power of a prime $p$ dividing $N !$ is $\sum_{j=1}^{\infty}\left\lfloor\frac{N}{p^{j}}\right\rfloor$. By considering the prime factorisation of $\left(\begin{array}{c}2 n \\ n\end{array}\right)$, show that

$$\frac{4^{n}}{2 n+1} \leqslant\left(\begin{array}{c} 2 n \\ n \end{array}\right) \leqslant(2 n)^{\pi(2 n)}$$

Setting $n=\left\lfloor\frac{x}{2}\right\rfloor$, deduce that for $x$ sufficiently large

$$\pi(x)>\frac{\left\lfloor\frac{x}{2}\right\rfloor \log 3}{\log x}>\frac{x}{2 \log x}$$