Proving that $\Gamma \left(n+ \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n}n!}$.

We can also exploit the fact that $\Gamma(x+1)=x\Gamma(x)$ to obtain \begin{align*} \Gamma\left(n + \frac{1}{2}\right) & =\left(n-1+\frac{1}{2}\right)\Gamma\left(n-1+\frac{1}{2}\right) \\ & =\left(n-1+\frac{1}{2}\right)\left(n-2+\frac{1}{2}\right)\Gamma\left(n-2+\frac{1}{2}\right) \\ & = \ldots = \left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\dots \ \frac{1}{2}\ \Gamma\left(\frac{1}{2}\right) \\ & = \frac{(2n-1)(2n-3)\cdots1}{2^n} \cdot \Gamma\left( \frac{1}{2}\right) \\ & =\frac{(2n-1)(2n-2)(2n-3)(2n-4)\cdots1}{2^n(2n-2)(2n-4)\dots2} \cdot\Gamma\left( \frac{1}{2}\right) \\ & =\frac{(2n-1)!}{2^{2n-1}(n-1)!}\Gamma\left( \frac{1}{2}\right)\\ & =\frac{2n!}{2^{2n}n!}\sqrt{\pi}. \end{align*}

An alternative way to prove the cited identity would be to use the integral representation of the $\Gamma$-function with arbitrary arguments :

$$ \Gamma(z)=\int_0^{\infty}e^{-t}t^{z-1} $$

Specifying $z=n+\frac{1}{2}$ gives us $$ \Gamma(n+\frac{1}{2})=\int_0^{\infty}e^{-t}t^{n-\frac{1}{2}} $$

Now choosing $t=x^2$ this transforms to

$$ \Gamma(n+\frac{1}{2})=2\underbrace{\int_0^{\infty}e^{-x^2}x^{2n}dx}_{I_n} $$

It is easy to show that $I_n=(-1)^n\frac{d^n}{da^n}\int_0^{\infty}e^{-a x^2}\big|_{a=0}=(-1)^n\frac{d^n}{da^n}\sqrt{\frac{\pi}{4a}}\big|_{a=0}$ where we have used the standard Gaussian integral in the last step.

Furthermore $\frac{d^n}{da^n}\sqrt{\frac{1}{a}}=\frac{(-1)^n}{2^n}(2n-1)!!=\frac{(-1)^n}{2^{2n}}\frac{2n!}{n!}$.

Putting everything together, we obtain

$$ \Gamma(n+\frac{1}{2})=\frac{\sqrt{\pi}}{2^{2n}}\frac{2n!}{n!} $$

as desired.