I saw this statement : $$\binom{2n}{n} \asymp \Theta \bigg(\frac{2^{2n}}{\sqrt{n}}\bigg) \asymp \Theta\bigg(\frac{4^n}{\sqrt{n}}\bigg)$$
How did we go from the first statement to the second? I tried Stirling's aproximation, but that didn't get me anywhere.
On
If we set: $$ I_n = \int_{0}^{\pi/2}\sin(x)^n\,dx \tag{1}$$ integration by parts gives: $$ \forall n\geq 2,\qquad I_n = \left(1-\frac{1}{n}\right) I_{n-2} \tag{2}$$ hence, by induction: $$ I_{2n}=\frac{\pi}{2}\binom{2n}{n}\frac{1}{4^n},\qquad I_{2n-1}=\frac{1}{2n}\left(\binom{2n}{n}\frac{1}{4^n}\right)^{-1}\tag{3}$$ and since $\{I_n\}_{n\geq 0}$ is trivially a decreasing sequence from $(1)$, $I_{2n}\leq I_{2n-1}$ gives: $$ \binom{2n}{n}\frac{1}{4^n}\leq\frac{1}{\sqrt{\pi n}}\tag{4} $$ that together with the consequences of $I_{2n}\geq I_{2n+1}$ proves: $$ \binom{2n}{n}\frac{1}{4^n}=\Theta\!\left(n^{-1/2}\right)\tag{5} $$ as wanted. We avoided Stirling's approximation.
If we want to avoid integrals, too, we may consider that: $$ \frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{6} $$ hence: $$\begin{align*}\left(\frac{1}{4^n}\binom{2n}{n}\right)^2&=\frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}\right)\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\\&=\frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{7}\end{align*}$$ but the product $$ \prod_{k=2}^{+\infty}\left(1+\frac{1}{4k(k-1)}\right) $$ is converging since the series $$ \sum_{k\geq 2}\frac{1}{4k(k-1)}=\frac{1}{4}\sum_{k\geq 2}\left(\frac{1}{k-1}-\frac{1}{k}\right)=\frac{1}{4}$$ is converging, proving $(5)$, again.
Stirling's approximation gives $$\binom{2n}{n} = \frac{(2n)!}{n!^2} \sim \frac{\sqrt{2\pi \cdot 2n} \left( \frac{2n}{e} \right)^{2n}}{\left[\sqrt{2\pi n} \left( \frac{n}{e} \right)^n \right]^2} = \frac{2\sqrt{\pi n} \cdot 2^{2n} \cdot \left( \frac{n}{e} \right)^{2n}}{2 \pi n \cdot \left( \frac{n}{e} \right)^{2n}} = \frac{1}{\sqrt{\pi n}} 2^{2n}$$