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.

  • 1
    $\begingroup$ Which is bigger, $n$ or $m$? $\endgroup$ Mar 4, 2013 at 0:14
  • $\begingroup$ Are there any restrictions on $m$ and $n$? $\endgroup$
    – Daryl
    Mar 4, 2013 at 0:15
  • $\begingroup$ Write them in term of one variable, e.g. let $m=n+a$ where $a \geq 0$ $\endgroup$
    – jimjim
    Mar 4, 2013 at 0:15
  • $\begingroup$ The are positive integers. In the particular application $m=n+1$. $\endgroup$
    – Maesumi
    Mar 4, 2013 at 0:16
  • $\begingroup$ @Jean-Sébastien text of question was corrected. $\endgroup$
    – Maesumi
    Mar 4, 2013 at 0:22

2 Answers 2


Consider $m = n+1$. Then $(m!)^n > (n!)^m$.


$$ \begin{eqnarray*} (m!)^n &=& ([n+1]!)^n \\ &=& (n+1)^n \cdot (n!)^n \\ &=& \frac{(n+1)^n}{ (n!) } (n!)^{n+1}\\ &=& \frac{(n+1)^n}{ (n!) } (n!)^{m}\\ \end{eqnarray*} $$ Now since $n+1 > k$ for all $k = 1, 2, \ldots, n$, It is clear that $\frac{(n+1)^n}{n!} > 1$, proving the result.

A simple induction then shows if $m > n$, then $(m!)^n > (n!)^m$ in general.


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

$$(n!)^{m}\lt (m)!^n$$


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