I've this expression to simplify:
$$\frac {1}{2^{3N}}\frac{(2m\pi)^{3N/2}}{h^{3N}\sqrt{\frac{2\pi3N}{2}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}}U^{3N/2}\frac{V^N}{\pi^{3N}}$$
and I'm supposed to end with:
$$V^N\left(\frac{U}{N}\right)^{3N/2}\left(\frac{me}{3h^2\pi}\right)^{3N/2}$$
But whenever I try I couldn't reach it
We have: $$I = \frac {1}{2^{3N}}\frac{(2m\pi)^{3N/2}}{h^{3N}\color{red}{\sqrt{\frac{2\pi3N}{2}}}\left(\frac{3N}{2}\right)^{3N/2}e^{-3N/2}}U^{3N/2}\frac{V^N}{\pi^{3N}}$$
$$I = \frac{2^{3N/2}m^{3N/2}\pi^{3N/2}e^{3N/2}U^{3N/2}V^N\times 2^{3N/2}}{2^{3N}h^{3N}3^{3N/2}N^{3N/2}\pi^{3N}}$$
$$I= \frac{2^{3N}}{2^{3N}} \times \frac{\pi^{3N/2}}{\pi^{3N}}\times \frac{(me)^{3N/2}}{3^{3N/2}h^{\color{green}{2}\times 3N/2}N^{3N/2}} \times U^{3N/2}\times V^N$$
Simplify it and we are done.
Unfortunately, I have to say that though the Stirling approximation to $\Gamma(3N/2+1)$ is correct, I never used that term: $\sqrt{\dfrac{2\pi3N}2}$, but I still get the required answer. Maybe, you just give the original expression without any simplification and we will be able to deduce from it.