If $m>n$, is $(m!)^n or\ (n!)^m$?

Question

m and n are two positive numbers and m>n. Which one is greater? $(m!)^n or\ (n!)^m$

I solved this question when m and n are natural numbers. I am wondering if there is any suggestion for the question when m and n are positive real numbers.

Answer 1

Everywhere below $z$ is assumed to be a positive real number.

Let start with identity: $$ \log\Gamma(1+z)=\log z+\log\Gamma(z) =-\gamma z+\sum_{k\ge1}\left[\frac zk-\log\left(1+\frac zk\right)\right], $$ where $\gamma$ is the Euler–Mascheroni constant.

Then: $$\Phi(z):=\frac{\log\Gamma(1+z)}z= -\gamma+\sum_{k\ge1}\left[\frac 1k-\frac1z\log\left(1+\frac zk\right)\right].\tag1 $$

Differentiating the equation $(1)$ over $z$ one obtains: $$ \frac{d\Phi(z)}{dz} =\frac1{z^2}\sum_{k\ge1}\left[\log\left(1+\frac zk\right)-\frac z{z+k}\right]>0.\tag2 $$ (The proof that all summands are positive is left as a simple exercise.)

From $(2)$ one concludes: $$z_1<z_2\implies\Phi(z_1)<\Phi(z_2) $$ which in turn means: $$ \frac{\log\Gamma(1+z_1)}{z_1}<\frac{\log\Gamma(1+z_2)}{z_2} \implies z_2\log\Gamma(1+z_1)<z_1\log\Gamma(1+z_2)\\ \implies (z_1!)^{z_2}<(z_2!)^{z_1}. $$

Answer 2

Partial answer

By Stirling's approximation

$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n $$

$$\implies (n!)^m \sim \sqrt{(2 \pi n)^m}\left(\frac{n}{e}\right)^{nm} \quad (m!)^n=\sqrt{(2 \pi m)^n}\left(\frac{m}{e}\right)^{nm}$$

and

$$\frac{(m!)^n}{(n!)^m}=\frac1{(2\pi)^{m-n}}\left(\frac{m^n}{n^m}\right)\left(\frac{m}{n}\right)^{nm}=\frac1{(2\pi n)^{m-n}}\left(\frac{m}{n}\right)^{n(m+1)}$$

and by $m=n+k$

$$\frac{(m!)^n}{(n!)^m}=\frac1{(2\pi n)^{k}}\left(1+\frac k n\right)^{n(n+k+1)}$$

which seems to have a minimum greater than $1$ for $k=1$.