Show that $n^{\pi\left(2n\right)-\pi\left(n\right)}<2^{2n}$ and $2^n\le\left(2n\right)^{\pi(2n)}$ for all $n>2$

There is a classical approach without much computation.

For any prime $p$, we may estimate its exponent in $\binom{2n}{n}$ as follows $$\nu_p\biggl(\binom{2n}{n}\biggr)=\sum_{k=1}^{\lfloor\log_p2n\rfloor}\biggl(\Bigl\lfloor\frac{2n}{p^k}\Bigr\rfloor-2\Bigl\lfloor\frac{n}{p^k}\Bigr\rfloor\biggr)\leqslant\sum_{k=1}^{\lfloor\log_p2n\rfloor}1\leqslant\log_p2n.$$ So we have $$\binom{2n}{n}=\prod_{\text{prime}~p\\p<2n}p^{\nu_p\bigl(\binom{2n}{n}\bigr)}\leqslant\prod_{\text{prime}~p\\p<2n}p^{\log_p2n}=\prod_{\text{prime}~p\\p<2n}2n=(2n)^{\pi(2n)}.$$ For $n>1$, we also have $$\binom{2n}{n}=\prod_{k=1}^n\frac{n+k}{k}>\prod_{k=1}^n2=2^n.$$ This proved the second inequality.

For the first one, just note that $\binom{2n}{n}$ is divisible by any prime $p$ in the region $(n,2n)$, therefore $$\binom{2n}{n}\geqslant\prod_{\text{prime}~p\\n<p<2n}p\geqslant\prod_{\text{prime}~p\\n<p<2n}n=n^{\pi(2n)-\pi(n)}.$$ Then $\binom{2n}{n}<\sum_{k=0}^{2n}\binom{2n}{k}=2^{2n}$ gives the disired result.

Using this result (and some discussions here)

$$\frac{x}{\ln{x}}<\pi(x)< 1.25506\frac{x}{\ln{x}} \text{, for } x \geq 17 \tag{1}$$

The second one $2^n \leq (2n)^{\pi(2n)}$ is a direct result of $\pi(x)>\frac{x}{\ln{x}}$ or $$\pi(2n)>\frac{2n}{\ln{(2n)}}>\frac{n\ln{2}}{\ln{(2n)}} \Rightarrow \pi(2n) \cdot \ln{(2n)} > n \cdot \ln{2} \Rightarrow \\ \ln{(2n)^{\pi(2n)}}>\ln{2^n} \Rightarrow (2n)^{\pi(2n)} > 2^n$$

The first one $$n^{\pi(2n)-\pi(n)}< 2^{2n} \Leftrightarrow \left(\pi(2n)-\pi(n) \right) \ln{n} < 2n \ln{2} \tag{2}$$ but $$1.25506 < 2\ln{2}$$ and it is enough to prove $$\color{red}{\pi(2n)-\pi(n) < \pi(n)} < 1.25506\frac{n}{\ln{n}} < 2 \ln{2}\frac{n}{\ln{n}}$$

which is covered here.