This question is from this following derivation from pages 204-205 of PDE Evans, 2nd edition. This is from the same proof of Example 9 on those pages of the textbook, and I asked a question about that as well (see Compact support).
Now $n=2k+1$ and $\alpha(n) = \frac{\pi^{n/2}}{\Gamma(\frac n2+1)} = \frac{\pi^{k+\frac 12}}{\Gamma(\frac n2+1)}$. Since $\Gamma\left(\frac 12\right)=\pi^{1/2}$ and $\Gamma(x+1)=x\Gamma(x)$ for $x > 0$, we can compute $$\frac{n \alpha(n)}{\pi^k 2^{k+1}}=\frac{n\pi^{1/2}}{2^{k+1}\Gamma(\frac n2+1)}\color{blue}{=\frac{1}{(n-2)(n-4)\cdots5\cdot3}=\frac{1}{\gamma_n}}$$
May I ask how were the last two equalities obtained (highlighted in $\color{blue}{\textbf{blue}}$)? All I know so far was that the first equality is due to plugging in the expression of $\alpha(n) = \frac{\pi^{k+\frac 12}}{\Gamma(\frac n2+1)}$.
The functional equation $\Gamma(x+1) = x\Gamma(x)$ yields
$$\begin{align} \Gamma\left(\tfrac{n}{2}+1\right) &= \Gamma\left(\tfrac{n}{2}\right)\tfrac{n}{2}\\ &= \Gamma\left(\tfrac{n}{2}-1\right)\tfrac{n}{2}\tfrac{n-2}{2}\\ &= \Gamma\left(\tfrac{n}{2}-r\right)\prod_{m=0}^r\tfrac{n-2m}{2} \\ &= \Gamma\left(\tfrac{1}{2}\right)\cdot\prod_{m=0}^{k}\left(\tfrac{n}{2}-m\right)\\ &= \pi^{1/2}\prod_{m=0}^k\frac{n-2m}{2}\\ &= \frac{n\pi^{1/2}}{2^{k+1}}\prod_{m=1}^k(n-2m). \end{align}$$
Plug that in, and you have the first of the equalities. For the second, the definition of $\gamma_n$ would probably be helpful.