if f(1) + f(2) +... + f(n-1)< the integral from 1 to n of f(x)dx < f(2) + f(3) +...+ f(n).
Chose f(x) = ln x. Show that n^n/e^(n-1)< n! < n^(n+1)/e^n
Assume that $\displaystyle (n!)'>0$ and $\displaystyle n>0$
$\displaystyle \begin{aligned}\sum_{k=1}^{n-1}\ln(k)&=\ln(1)+\ln(2)+\cdots+\ln(n-1)\\
&=\ln\left(1\cdot{2}\cdots{n-1}\right)\\
&=\ln((n-1)!)\end{aligned}$
And
$\displaystyle \int_1^n\ln(x)dx=n\ln(n)-n+1=\ln\left(\frac{n^n}{e^{n-1}}\right)$
As well as
$\displaystyle \begin{aligned}\sum_{k=2}^{n}\ln(k)&=\ln(2)+\ln(3) +\cdots+\ln(n)\\
&=\ln\left(1\cdot{2}\cdots{n}\right)\\
&=\ln(n!)\end{aligned}$
So we would have that
$\displaystyle \ln((n-1)!)<\ln\left(\frac{n^n}{e^{n-1}}\right)<\ln(n!)$
And since the exponential function is stricly increasing across its domain the above inequality also gives
$\displaystyle (n-1)!<\frac{n^n}{e^{n-1}}<n!$
So then because $\displaystyle n!$ is strictly increasing we may rewrite our inequality as
$\displaystyle 1<\frac{n^n}{e^{n-1}}\cdot\frac{1}{(n-1)!}<n$
Reciporcating and not forgetting to reverse the direction of the inequalities gives
$\displaystyle \frac{1}{n}<(n-1)!\cdot\frac{e^{n-1}}{n^n}<1$
Multiplying through by $\displaystyle \frac{n^n}{e^{n-1}}$ gives
$\displaystyle \frac{n^{n-1}}{e^{n-1}}<(n-1)!<\frac{n^n}{e^{n-1}}$
Finally multiplying through by $\displaystyle n$ gives
$\displaystyle \frac{n^n}{e^{n-1}}<n!<\frac{n^{n+1}}{e^{n-1}}\quad\blacksquare$