Show that $n^{n!} \gt (n!)^{n}$ for $n\gt 2 \in \mathbb{N}$

I have tried induction taking the base case $n=3$
It is not going smooth however

I am looking for some simpler proof, induction or contradiction or anything is fine. Thanks!

  • 2
    $\begingroup$ In general, for $3\leq m<n$ we have $m^n > n^m$. Maybe that's easier to prove? $\endgroup$ – Arthur Oct 29 '14 at 15:57
  • $\begingroup$ Oh right! thanks a lot I'll try xD $\endgroup$ – pooja Oct 29 '14 at 16:01
  • $\begingroup$ try it by log $$n! log n > n log(n!) $$ $\endgroup$ – Khosrotash Oct 29 '14 at 16:05

$$n^{n!} = \underbrace{n \times n \times \cdots \times n}_{n! \text{ terms}} \tag 1$$

$$(n!)^n = \underbrace{n! \times \cdots \times n!}_{n \text{ terms}} = \underbrace{1 \times \cdots \times n \times 1 \times \cdots \times n \times \cdots 1\times \cdots \times n}_{n^2 \text{ terms}} \tag 2$$

$n! > n^2$ when $n \geq 4$, and each term in $(1)$ is bigger than terms in $(2)$.

Now verify for $n=3$ by a bit of computation

  • $\begingroup$ brilliantly done! thank you so much XD $\endgroup$ – pooja Oct 29 '14 at 16:45

Hint: $a^{\frac1a}$ is strictly decreasing for $a> e$.
Indeed, its derivative is $(\exp(\frac1a\log a))'=a^{\frac1a}\cdot(\frac{-1}{a^2}\log a+\frac1a\frac1a)=a^{\frac1a-2}\cdot(1-\log a)$.


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