Factorial to exponential conversion

Question

I'm solving a statistical mechanics problem and in the solution, they have directly replaced

$$\left (\dfrac{Nd}{2}\right)! =\left (\dfrac{Nd}{2e}\right)^{Nd/2}$$

Can someone tell me how to derive this equality?

Answer 1

If $N$ is large, they may have used Stirling's approximation: $$n! \sim \sqrt{2\pi n}\left(\dfrac ne\right)^n$$

but I have no clue why the factor $\sqrt{\pi Nd}$ was omitted.

Answer 2

It's only an approximation, not a true equation. But it uses the Stirling approximation, $n!\approx \sqrt{2\pi n}(n/e)^n$. In this case with $n:=Nd/2$, the $\sqrt{2\pi n}$ factor has been neglected. In statistical mechanics it's often more important to consider the factorial's logarithm, because entropy is proportional to the logarithm of the number of microstates per macrostate. Asymptotically, the neglected term in the logarithm is negligible. Thus$$\ln n!\approx\tfrac12\ln(2\pi n)+n\ln(n/e)\sim n\ln(n/e)\sim n\ln n.$$