How to prove? $0<\left(n^2-1\right)^n(n+1)$$
\displaystyle 0<\left(n^2-1\right)^n(n+1)<n^{2n+1}
$$

How does one prove this inequality for $n>1\in \mathbb{N}$ ?

 A: The LHS is immediate since all the terms are positive. 
The RHS follows from AM-GM. We have
$$ (n^2-1)^n (n^2+n) < \left( \frac{n \times (n^2 -1) +1 \times  (n^2+n)}{n+1} \right) ^{n+1} =  (n^2)^{n+1}. $$
We have strict inequality since not all the terms are equal. 
Cancel 1 factor of $n$ from both sides.
A: $(n^2-1)^n\cdot (n+1) = (n-1)^n\cdot (n+1)^{n+1} < n^{2n+1} \iff \left(1-\dfrac{1}{n}\right)^n\cdot \left(1+\dfrac{1}{n}\right)^{n+1} < 1 \iff n\cdot ln\left(1-\dfrac{1}{n}\right) + (n+1)\cdot ln\left(1 + \dfrac{1}{n}\right) < 0 \iff \dfrac{ln(1-\dfrac{1}{n})}{\dfrac{1}{n}} + ln(1 + \dfrac{1}{n}) + \dfrac{ln(1 + \dfrac{1}{n})}{\dfrac{1}{n}} < 0$.
Let $x = \dfrac{1}{n}$, then $0 < x < 1$, and the inequality to be proven is:
$\dfrac{ln(1-x^2)}{x} + ln(1 + x) < 0 \iff \dfrac{ln(1-x^2) + xln(1+x)}{x} < 0$. To complete the proof, we show: $f(x) < 0$ with $f(x) = ln(1-x^2) + xln(1+x)$ on $(0,1)$. Thus:
$f'(x) = -\dfrac{2x}{1-x^2} + ln(1+x) + \dfrac{x}{1+x} = ln(1+x) - \dfrac{x}{1-x} = g(x)$. Observe that if $f'(x) < 0$, then $f(x) < f(0) = 0$, and we're done. So we need to show again that: $f'(x) = g(x) < 0$ on $(0,1)$.
To get this, differentiate $g$ and have:
$g'(x) = \left(ln(1+x) + 1 + \dfrac{1}{x-1}\right)' = \dfrac{1}{1+x} - \dfrac{1}{(1-x)^2} =\dfrac{x(x-3)}{(1+x)(1-x)^2} < 0$. Thus: $g(x) < g(0) = 0$, and we're done.
