How to lower/upper bound $n!$ using $1+x\leq e^x$? I need to prove for all positive integer $n$
$$
e\left(\frac{n}{e}\right)^n\leq n!\leq en\left(\frac{n}{e}\right)^n,
$$ 
using the hint $1+x\leq e^x$ for all $x\in \mathbb{R}$.
I did this:
The hint says 


*

*for $x=0$, $1\leq 1$;

*for $x=1$, $2\leq e$;

*...

*for $x=n-1$, $n\leq e^{n-1}$.


So I multiplied these $n$ inequalities to get
$$
n!\leq e^{n-1+\ldots+1},
$$
or 
$$
n!\leq e^{\frac{n(n-1)}{2}},
$$
 and I get stuck there. 
 A: Rearranging gives the equivalent inequalities
$$1 \le (n-1)! \left(\frac{e}{n}\right)^{n-1} \le n.$$
When $n=1$ both inequalities are equalities.
Assuming the statement holds for $n=k$, then we want to prove
$$1 \le k! \left(\frac{e}{k+1}\right)^k \le k+1.$$

The first inequality holds since
$$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\ge 1} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\ge 1 \text{ by induction}}$$
where the first term is $\ge 1$ because the hint gives
$$e^{1/k} \frac{k}{k+1} \ge \left(1+\frac{1}{k}\right) \frac{k}{k+1} = 1.$$

The second inequality is due to
$$k! \left(\frac{e}{k+1}\right)^k = \underbrace{e \cdot \left(\frac{k}{k+1}\right)^k}_{\le \frac{k+1}{k}} \cdot \underbrace{(k-1)! \left(\frac{e}{k}\right)^{k-1}}_{\le k \text{ by induction}}$$
where the first term is $\le \frac{k+1}{k}$ because
$$e \cdot \left(\frac{k}{k+1}\right)^{k+1}
= e \cdot \left(1-\frac{1}{k+1}\right)^{k+1} \le e \cdot e^{-1} = 1.$$
[It is well known that $\left(1-\frac{1}{k+1}\right)^{k+1}$ converges to $e^{-1}$ as $k \to \infty$, but one can show by induction that this actually increases monotonically to $e^{-1}$.]
A: Take the log of all sides, so we just need to prove that
$$n\ln n-n+1\le\ln(n!)\le (n+1)\ln n-n+1$$
Then, it is easy to see that
$$\ln(n!)=\sum_{k=1}^n\ln(k)$$
And that is the right Riemann sum to the following integral:
$$\sum_{k=1}^n\ln(k)\ge\int_1^n\ln(x)\ dx=n\ln n-n+1$$
So we have proven the left side.  Then notice that we have the following trapezoidal sum:
$$\frac{\ln(n)+2\ln((n-1)!)}2=\frac{\ln(n!)+\ln((n-1)!)}2\\=\sum_{k=1}^n\frac{\ln(k+1)+\ln(k)}2\\\le\int_1^n\ln(x)\ dx=n\ln(n)-n+1\\\frac{\ln(n)+2\ln((n-1)!)}2\le n\ln(n)-n+1\\\implies\ln((n-1)!)\le\left(n-\frac12\right)\ln(n)-n+1\\\implies\ln(n!)=\ln(n)+\ln((n-1)!)\le\left(n+\frac12\right)\ln(n)-n+1$$
Thus, when $n\ge1$,
$$\ln(n!)\le\left(n+\frac12\right)\ln(n)-n+1\le(n+1)\ln(n)-n+1$$
and we have everything we wanted.
