TITLE: Prove inequality $n^{n^2-1} \leq e^{n^n-n}$ for $n = 4, 6, 8, 10, \dots$
QUESTION [5 upvotes]: Prove 
  $\,n^{n^2-1} < e^{n^n-n}$ for $n = 4, 6, 8, 10, \dots$

or

Prove $\,\dfrac {n^{n^2}}{e^{n^n-n}} <n$ for $n=4, 6, 8, 10, \dots$

I'm struggling with some hard problems,
and if I could prove this inequality, I can make a result.
Is this inequality true?
(I have already used WolframAlpha for checking some $n$, so I think this is true.)
Thanks a lot.

REPLY [5 votes]: Taking the logarithm we get the equivalent inequality
$$(n^2-1)\ln(n)+n \leq n^n=e^{n\ln(n)}.$$
Now for $x\geq 0$, consider the inequality $e^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}\geq x+\frac{x^3}{3!}$. Then, for $n\geq 4$,
$$e^{n\ln(n)}\geq \underbrace{n\ln(n)}_{\geq n}+\underbrace{\frac{n^3\ln^3(n)}{6}}_{\geq (n^2-1)\ln(n)}\geq n+ (n^2-1)\ln(n)$$