Prove that $\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}For all $n\in\mathbb N$ prove that:
$$\frac{(n+2)^{n+1}}{(n+1)^n}-\frac{(n+1)^n}{n^{n-1}}<e$$
We can rewrite it in the following form 
$$(n+2)\left(1+\frac{1}{n+1}\right)^{n}-(n+1)\left(1+\frac{1}{n}\right)^{n-1}<e$$ or
$$(n+1)\left(1+\frac{1}{n+1}\right)^{n+1}-n\left(1+\frac{1}{n}\right)^{n}<e$$
and how should I proceed?
 A: I do not know if this is a satisfactory answer for you.
Consider $$A=\frac{(n+2)^{n+1}}{(n+1)^n}\qquad , \qquad B=\frac{(n+1)^n}{n^{n-1}}$$ Take logarithms and expand as Taylor series for large values of $n$. This would lead to $$\log(A)=1+\log \left(n\right)+\frac{1}{2 n}+\frac{1}{3
   n^2}-\frac{13}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\log(B)=1+\log \left(n\right)-\frac{1}{2 n}+\frac{1}{3
   n^2}-\frac{1}{4 n^3}+O\left(\frac{1}{n^4}\right)$$ Taylor again $$A=e^{\log(A)}=e n+\frac{e}{2}+\frac{11 e}{24 n}-\frac{43 e}{48
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$B=e^{\log(B)}=e n-\frac{e}{2}+\frac{11 e}{24 n}-\frac{7 e}{16
   n^2}+O\left(\frac{1}{n^3}\right)$$ $$A-B=e-\frac{11 e}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
A: Claude Leibovici's answer prove only that the proposition is true for sufficiently large rather than arbitrary positive $n$. Here is the complete proof with stronger result.

Let 
$$f(x):=x\left(1+\frac{1}{x}\right)^{x},\ x>0.$$ 
$$f'(x) = e^{x\ln(1+\frac1x)}\Big[1+x\Big(\ln\Big(1+\frac1x\Big)-\frac1{1+x}\Big)\Big].$$
$f$ increases since
$$\ln\Big(1+\frac1x\Big)=-\ln\Big(1-\frac1{1+x}\Big)>\frac1{1+x}$$
by Taylor expansion of the logarithmic function around $1$. Because $f(x)$ is strictly convex, $1=f'(0)<f'(x)<f'(\infty)=e,\ \forall x\in (0,\infty)$.
So we conclude more than the required inequality
$$1<f(n)-f(n-1)<f(n+1)-f(n) = \int_n^{n+1}f'(x)dx<e,\ \forall n\ge 1.$$
