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.
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1Which is bigger, $n$ or $m$? – Jean-Sébastien Mar 04 '13 at 00:14
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Are there any restrictions on $m$ and $n$? – Daryl Mar 04 '13 at 00:15
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Write them in term of one variable, e.g. let $m=n+a$ where $a \geq 0$ – jimjim Mar 04 '13 at 00:15
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The are positive integers. In the particular application $m=n+1$. – Maesumi Mar 04 '13 at 00:16
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@Jean-Sébastien text of question was corrected. – Maesumi Mar 04 '13 at 00:22
2 Answers
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Consider $m = n+1$. Then $(m!)^n > (n!)^m$.
Proof:
$$ \begin{eqnarray*} (m!)^n &=& ([n+1]!)^n \\ &=& (n+1)^n \cdot (n!)^n \\ &=& \frac{(n+1)^n}{ (n!) } (n!)^{n+1}\\ &=& \frac{(n+1)^n}{ (n!) } (n!)^{m}\\ \end{eqnarray*} $$ 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.
Shaun Ault
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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
$$(n!)^{m}\lt (m)!^n$$
Chris Brooks
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