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.
 A: 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.
A: 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.
A: I'll enhance an earlier answer of mine 
and show that
$e(n/e)^n < n! < e^{3/2}\sqrt{n}(n/e)^n$.
This is better than OP's request
since
$((n+1)/e)^n  < e(n/e)^n$
is the same as
$(1+1/n)^n < e$
and
$e^{3/2}\sqrt{n}(n/e)^n
<e((n+1)/e)^{n+1}$
is the same as
$e\sqrt{e/n}<(1+1/n)^{n+1}$
which is more than true since
$(1+1/n)^{n+1} > e$
(proofs available upon request).
The only result from analysis we need is
 $z-z^2/2 < \ln(1+z) < z$
for $0 < z < 1$.
We first bound
$H_n = \sum_{k=1}^{n} 1/k$.
$\begin{align}
H_n &= \sum_{k=1}^{n} 1/k\\
&> \sum_{k=1}^{n} (\ln(1+1/k))\\
&= \sum_{k=1}^{n} (\ln(k+1)-\ln(k))\\
&= \ln(n+1)\\
\end{align}
$
and
$\begin{align}
H_n &= \sum_{k=1}^{n} 1/k\\
&< \sum_{k=1}^{n} (\ln(1+1/k) -1/(2k^2))\\
&= \sum_{k=1}^{n} (\ln(k+1)-\ln(k)) - (1/2)\sum_{k=1}^{n} 1/k^2\\
&= \ln(n+1)- (1/2)\sum_{k=1}^{n} 1/k^2\\
\end{align}
$.
To bound the right-hand sum
$\begin{align}
\sum_{k=1}^{n} 1/k^2
&=1+\sum_{k=2}^{n} 1/k^2\\
&<1+\sum_{k=2}^{n} 1/(k(k-1))\\
&=1+\sum_{k=2}^{n} (1/(k-1)-1/k)\\
&=1+1-1/n\\
&< 2\\
\end{align}
$
so
$\ln(n+1)< H_n< \ln(n+1)+1$.
We start with
$\ln(n!) =
\sum_{k=1}^n \ln k$
and estimate $\ln k$.
$\begin{align}
(x+1)\ln(x+1)-x \ln x
&= x(\ln(x+1)-\ln(x))+\ln(x+1)\\
&=x\ln(1+1/x)+\ln(x+1)\\
\end{align}
$
so $ (x+1)\ln(x+1)-x \ln x -\ln(x+1)=x\ln(1+1/x)$.
Using this,
$x\ln(1+1/x)
<x(1/x) = 1
$
and
$x\ln(1+1/x)
>x(1/x-1/(2x^2)
=1-1/(2x)
$
so
$1-1/(2x) < (x+1)\ln(x+1)-x \ln x -\ln(x+1) < 1$.
Note that this is just an approximate form of
$\int \ln x\,dx = x \ln x - x$
or $\ln x = (x \ln x)' - 1$.
Summing for $x$ from 1 to $n-1$,
$n-1-\sum_{k=1}^{n-1} 1/(2k) < \sum_{x=1}^{n-1}  
\big((x+1)\ln(x+1)-x \ln x -\ln(x+1)\big) < n-1
$
or
$n-1-H_{n-1}/2 < \sum_{x=1}^{n-1}  
\big((x+1)\ln(x+1)-x \ln x -\ln(x+1)\big) < n-1
$
Since the left part of the sum is telescoping
and the right part gives $\ln(n!)$,
$n-1-H_{n-1}/2 < n \ln n -\ln(n!) < n-1$.
Using the right side,
$\ln(n!) > n \ln n-n+1$
or
$n! > e(n/e)^n$.
Using the left side,
$\ln(n!) 
< n \ln n-n+1+H_{n-1}/2
< n \ln n-n+1+(\ln(n)+1)/2
$
or
$n! < e^{3/2}\sqrt{n}(n/e)^n$.
By taking more terms of 
the expansion of $\ln(1+1/z)$,
we can get the
asymptotic series for $\ln(n!)$,
which is Stirling's formula without the constant.
