<s>[INST] How to prove $(\frac{n+1}{e})^nHow to prove $$\left(\frac{n+1}{e}\right)^n<n!<e\left(\frac{n+1}{e}\right)^{n+1}$$ without integrating method? In fact we could prove this by noticing that 
$$i<x<i+1\Rightarrow \ln i<\ln x<\ln (i+1),$$ then integrating it.
 [/INST] Induct - the base case $n = 1$ is trivial. We will show both sides of the equality by looking at ratios; indeed, $\displaystyle\frac{(n+2)^{n+1} e^n}{e^{n+1} (n+1)^n} = \frac{1}{e}\cdot\left(1+\frac{1}{n+1}\right)^{n} \cdot(n+2) < \frac{1}{1+1/(n+1)} \cdot (n+2) = n+1$ since $\left(1 + \frac{1}{n+1}\right)^{n+1} < e.$ 
Hence, $\left(\frac{n+2}{e}\right)^{n+1} < \left(\frac{n+1}{e}\right)^{n}\cdot (n+1) < n!\cdot (n+1) = (n+1)!$ by hypothesis. For the other side, the ratio is $\displaystyle\frac{1}{e}\cdot \left(1 + \frac{1}{n+1}\right)^{n+1}\cdot(n+2)$; it suffices to show that this is greater than $n+1.$ That is, $\displaystyle \frac{1}{e}\cdot\left(\frac{n+2}{n+1}\right)^{n+1} > \frac{n+1}{n+2},$ which is equivalent to $\left(1+ \frac{1}{n+1}\right)^{n+2} > e.$ To prove this, we use the fact that $\ln (1+h) > \frac{h}{1+h}.$ Then $(n+2)\ln\left(1+\frac{1}{n+1}\right) > (n+2)\frac{1/(n+1)}{(n+2)/(n+1)} = 1,$ whence the desired inequality follows.
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