Seems Complicated, but is not really 1 !

$\begin{aligned} N & = \frac {2^{2n+1}}{\sqrt \pi (n+1)!} \int_0^\infty t^{n+\frac 12}e^{-t} dt \\ & = \frac {2^{2n+1}}{\sqrt \pi (n+1)!} \Gamma \left(n + \frac 32\right) & \small \blue{\text{where }\Gamma (\cdot) \text{ denotes the gamma function.}} \\ & = \frac {2^{2n+1}}{\sqrt \pi (n+1)!} \left(\frac {2n+1}2 \times \cdots \times \frac 52 \times \frac 32 \times \frac 12 \Gamma \left(\frac 12\right) \right) & \small \blue{\text{Since }\Gamma(1+s) = s\Gamma(s)} \\ & = \frac {2^{2n+1}(2n+1)!!\sqrt \pi}{\sqrt \pi (n+1)! 2^{n+1}} & \small \blue{\text{and }\Gamma \left(\frac 12 \right) = \sqrt \pi} \\ & = \frac {2^{2n+1}(2n+1)!}{(n+1)! 2^{n+1} 2^nn!} & \small \blue{\text{As }(2n+1) = \frac {(2n+1)!}{2^n n!}} \\ & = \frac {(2n+1)!}{n!(n+1)!} \\ & = \boxed{\binom {2n+1}n} \end{aligned}$

Reference: Gamma function

The value of the integral is $\Gamma \left(n+\dfrac{3}{2}\right)$ . The remaining is algebraic calculations.