Prove that $(n!)! > n[(n-1)!]^{n!}$ 
Prove that for a natural number $n > 2$, $$(n!)! > n[(n-1)!]^{n!}.$$

I thought about proving it by induction but was wondering if there was an easier way.
 A: Rewriting  the expression, you want to show that $$(n!)! > n \left(\frac{n!}{n}\right)^{n!}$$ or $$\frac{\left(\frac{n!}{n}\right)^{-n!} (n!)!}{n}>1\implies \log \left(\frac{\left(\frac{n!}{n}\right)^{-n!} (n!)!}{n}\right)>0$$ The only thing I have been able to do is a Taylor expansion around $n=2$; for the lhs, this gives $$2 (\gamma -1)^2 (n-2)+\left(-6+7 \gamma -\gamma ^3+\frac{1}{6} (\gamma -1) (2
   \gamma -5) \pi ^2\right) (n-2)^2+O\left((n-2)^3\right)$$ in which all coefficients are positive. Expanding to $O\left((n-2)^4\right)$, the coefficients becomes a monster (which I shall not write) which is again positive. Pushing the calculation to $O\left((n-2)^{10}\right)$ still show only positive coefficients  which are smaller and smaller.
But I am afraid that this does not prove anything.
I also tried $$u_n=\frac{((n-1)!)^{-n!} (n!)!}{n}$$ and used Taylor expansion for $$\log \left(\frac{u_{n+1}}{u_n}\right)=\log \left(\frac{15}{4}\right)+(n-2) \left(\frac{937}{60}-11 \log (2)+\gamma 
   \left(-\frac{157}{10}+4 \gamma +\log (64)\right)\right)+O\left((n-2)^2\right)$$ where again all terms are positive as well as in higher order expansions.
I have not been able to do anything for infinitely large values of $n$ using Stirling like formaulae.
May I confess that I am not happy at all with this answer ?
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\pars{n!}! > n\bracks{\pars{n - 1}!}^{\, n!}:\ ?.\qquad n = 3,4,5,\ldots}$

Note that it's equivalent to prove
$\ds{\quad\pars{n!}! > {\pars{n!}^{\, n!} \over n^{n! - 1}}\quad}$ or/and
$$
{m^{m} \over m!} < n^{m - 1}\quad\ \mbox{with}\quad m\ \equiv n!
$$.
For $\ds{\color{#f00}{n = 3}}$ it is obviously true because
$$
m = 3! = 6 \implies
\pars{~{m^{m} \over m!} = {6^{6} \over 6!} = {324 \over 5} < n^{m - 1} = 3^{5} = 243~}
$$

Then, we can concentrate in the case $\ds{\color{#f00}{n \geq 4}}$:
\begin{align}
{m^{m} \over m!} < \sum_{k = 0}^{\infty}{m^{k} \over k!} = \expo{m} =
\bracks{\expo{}\,\pars{\expo{} \over n}^{m - 1}}n^{m - 1}
\end{align}
The factor $\ds{\expo{}\,\pars{\expo{} \over n}^{m - 1} < 1}$ for
$\ds{\color{#f00}{n \geq 4}}$ which
$\underline{can\ be\ proved\ by\ induction}$:


*

*$\ds{\left.\expo{}\pars{\expo{} \over n}^{n! - 1}\right\vert_{\ n\ =\ 4} =
\expo{}\pars{\expo{} \over 4}^{23} \approx 3.7643 \times 10^{-4} < 1}$
$$\mbox{}$$

*$\ds{%
\expo{}\,\pars{\expo{} \over n + 1}^{\pars{n + 1}! - 1}  <
\expo{}\,\pars{\expo{} \over n}^{\pars{n + 1}! - 1} =
\pars{\expo{} \over n}^{n\pars{n!}}
\bracks{\expo{}\,\pars{\expo{} \over n}^{n! - 1}} \color{#f00}{< 1}}$


This finishes the proof. 

A: First note that for $m>2$:
$$m!>\left(\frac{m}{e}\right)^m\sqrt{2\pi m}$$
We first show that if $k>e$ and $n>2$ then
$$(n-1)\ln k>n-\frac{1}{2}\ln n-\ln\sqrt{2\pi}$$
which doesn't seem hard since $\ln\sqrt{2\pi}\approx0.92$. Now from this inequality we can see that:
$$n(\ln k-1)>\ln k-\ln\sqrt{2\pi n}\implies \left(\frac{k}{e}\right)^n>\frac{k}{\sqrt{2\pi n}}$$
So if $k>e$ then it is safe to say that
$$\left(\frac{m}{e}\right)^m\sqrt{2\pi m}>\left(\frac{m}{k}\right)^m k$$
for $m>2$. Back to the original question, let $n!=m$, you need to show
$$m!>\left(\frac{m}{n}\right)^m n$$
which shouldn't cause any trouble now.
