To evaluate the LHS introduce
$$f(z) = (-1)^n n! \frac{1}{z+n+1} \prod_{q=0}^n \frac{1}{z-q}.$$
This has the property that
$$\mathrm{Res}_{z=r} f(z)
= (-1)^n n! \frac{1}{r+n+1}
\prod_{q=0}^{r-1} \frac{1}{r-q}
\prod_{q=r+1}^n \frac{1}{r-q}
\\ = (-1)^n n! \frac{1}{r+n+1}
\frac{1}{r!}
\frac{(-1)^{n-r}}{(n-r)!}
= \frac{1}{r+n+1}
(-1)^r {n\choose r}.$$
As we seek to compute
$$\sum_{r=0}^n \frac{1}{r+n+1}
(-1)^r {n\choose r}$$
and residues sum to zero with the residue at infinity being zero by
inspection we get for our sum
$$-\mathrm{Res}_{z=-n-1} f(z)
= - (-1)^n n! \prod_{q=0}^n \frac{1}{-n-1-q}
= n! \prod_{q=0}^n \frac{1}{n+1+q}
\\ = \frac{n!\times n!}{(2n+1)!}.$$
On the other hand working with the RHS we have with the Legendre
duplication formula
$$\frac{\sqrt{\pi} 2^{-2n-1} n!}{(n+1/2)!}
= \frac{\sqrt{\pi} 2^{-2n-1} n!}{\Gamma(n+3/2)}
\\ = \frac{\sqrt{\pi} 2^{-2n-1} \times n! \times n!}
{\Gamma(n+1) \Gamma(n+3/2)}
\\ = \frac{\sqrt{\pi} 2^{-2n-1} \times n! \times n!}
{2^{1-2(n+1)} \sqrt{\pi} \Gamma(2n+2)}
= \frac{n!\times n!}{(2n+1)!}.$$
We see that the LHS and the RHS are identical as claimed.
