Prove inequality $n^{n^2-1} \leq e^{n^n-n}$ for $n = 4, 6, 8, 10, \dots$

Question

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.

Answer 1

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 $n\geq 4$, $$e^{n\ln(n)}\geq \underbrace{n\ln(n)}_{\geq n}+\underbrace{\frac{n^3\ln^3(n)}{6}}_{\geq (n^2-1)\ln(n)}\geq n+ (n^2-1)\ln(n)$$

Answer 2

Hint: taking the logarithm on both sides we obtain: $$(n^2-1)\ln(n)<n^n-n$$ so we get $$0<\frac{n^n-n}{n^2-1}-\ln(n)$$ Now define $$f(n)=\frac{n^n-n}{n^2-1}-\ln(n)$$ and now use calculus.