TITLE: Which is bigger $(n!)^m$ or $(m!)^n$?
QUESTION [8 upvotes]: Assume $m>n$, which is bigger $(m!)^n$ or  $(n!)^m$? 
This question came about during a Taylor series approximation.
Considering plot of $(n!)^{1/n}$ and Stirling's formula one guesses larger base wins.

REPLY [8 votes]: Repeated use of the fact $n!\leq n^n$ yields
$$n!\cdot n!\cdots n!\lt (n+1)^n\cdot (n+2)^n\cdots (n+i)^n$$ for any $i\gt 0$, where there are $i$ terms on the LHS. Then we have
$$(n!)^i\lt \Big( (n+1)\cdots(n+i)\Big)^n$$
or equivalently
$$(n!)^i\lt \left( \frac{(n+i)!}{n!}\right)^n.$$
Rearranging, this says
$$(n!)^{n+i}\lt (n+i)!^n$$
so if we know $m\gt n$ then $m=n+i$ for some $i\gt 0$, so this tells us finally that
$$(n!)^{m}\lt (m)!^n$$