How to find $\Gamma(N/2+1)$?

Question

I started exploring the gamma function and I stumbled upon this paper for my high school essay Volume and Surface Area of anN-Sphere. There it says that $\Gamma(\dfrac12)=\sqrt{\pi}$ and $\Gamma(x+1)=x\Gamma(x)$. But I can't quite catch why when finding $\Gamma(n/2+1)$, we employ the $\Gamma(1/2+1)$. $N$ is not necessarily $1$, is it? So how did they get the final result?

Answer 1

Repeated application of the functional equation $\Gamma(x+1)=x\Gamma(x)$ yields $$ \Gamma \left( {\frac{n}{2} + 1} \right) = \frac{n}{2}\Gamma \left( {\frac{n}{2}} \right) = \frac{n}{2}\left( {\frac{n}{2} - 1} \right)\Gamma \left( {\frac{n}{2} - 1} \right) = \cdots = \frac{n}{2}\left( {\frac{n}{2} - 1} \right) \cdots \frac{1}{2}\Gamma \left( {\frac{1}{2}} \right) \\ = \frac{n}{2}\left( {\frac{n}{2} - 1} \right) \cdots \frac{1}{2}\sqrt \pi $$ ($n$ is odd a positive).

Answer 2

$$\Gamma(x+1) = x \Gamma(x) \implies \Gamma\left(\frac{3}{2}+1\right) = \frac{3}{2}\Gamma\left(\frac{1}{2} + 1\right) = \frac{3}{2}\times \frac{1}{2}\Gamma\left(\frac{1}{2}\right)$$