Noam Elkies notes that there is a quick proof of $$\binom{2n}{n} \leq \frac{4^n}{\sqrt{\pi n}}$$ by writing $$\binom{2n}{n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x dx$$ and bounding $\cos^2 x \leq e^{-x^2}$.
There is an equally good lower bound by a similar method: $$\int_{-\pi/2}^{\pi/2} \cos^{2n} x dx =\int_{-\pi/2}^{\pi/2} \frac{\sec^2 x dx}{(\tan^2 x+1)^{n+1}} = \int_{- \infty}^{\infty} \frac{du}{(1+u^2)^{n+1}} \geq \int_{- \infty}^{\infty} e^{-(n+1) u^2} du $$ so $$\binom{2n}{n} \geq \frac{4^n}{\sqrt{\pi (n+1)}}.$$ Here the inequality $\tfrac{1}{1+u^2} \geq e^{-u^2}$ follows from the standard bound $e^y \geq 1+y$ for $y\geq 0$.