TITLE: How to prove $(\frac{n+1}{e})^n<n!<e(\frac{n+1}{e})^{n+1}$ without integrating method?
QUESTION [7 upvotes]: How 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.

REPLY [2 votes]: The left hand inequality is immediate from the power series for $e^x$. Plugging in $x = n+1$ gives
$$e^{n+1} = \sum_{m=0}^{\infty} {(n+1)^m \over m!} > {(n+1)^{n+1} \over (n+1)!}$$
$$= {(n+1)^n \over n!}$$
This rearranges into the left inequality.