Prove $\left(\frac{n+1}{\text{e}}\right)^nHow to prove this inequality?
$$\left(\frac{n+1}{\text{e}}\right)^n<n!<\text{e}\left(\frac{n+1}{\text{e}}\right)^{n+1}$$
 A: Multiplying the whole thing through by $\,e^n\,$ and doing a little algebra we get an easier, imo, expression to prove:
$$(n+1)^n<e^nn!<(n+1)^{n+1}$$
$$n=1:\;\;\;\;\;\;2<e\cdot 1!<(2)^2\Longleftrightarrow 2<e<4\ldots\text{true.}$$
Now assume for $\,n\,$ and prove for $\,n+1\,$:
$$(n+1)^n<e^nn!<(n+1)^{n+1}\stackrel ?\Longrightarrow (n+2)^{n+1}<e^{n+1}(n+1)!<(n+2)^{n+2} $$
Let us focus on the right inequality:
$$e^{n+1}(n+1)!=e(n+1)e^nn!\stackrel{\text{ind. hypot.}}<e(n+1)(n+1)^{n+1}=e(n+1)^{n+2}$$
So it is enough now to prove that
$$e(n+1)^{n+2}<(n+2)^{n+2}\Longleftrightarrow e<\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{n+1}\right)\ldots$$
and I'll let the final touch to you (hint: remember the limit definition of the number $\,e\,$ ...)
For the left hand inequality: the same inductive assumption and
$$e^{n+1}(n+1)!=e(n+1)e^nn!\stackrel{\text{ind. hyp.}}>e(n+1)(n+1)^n=e(n+1)^{n+1}$$
so it's enough to show
$$e(n+1)^{n+1}>(n+2)^{n+1}\Longleftrightarrow e>\left(1+\frac{1}{n+1}\right)^{n+1}$$
Again, with the same hint as at the end of the first part, give the final touch here.
A: First denote, that $\log n! = \sum_{x=1}^n \log x$. Now we can simply bound the sum from above and below by:
\begin{align}
& \int_1^n \log  (x) \, dx \leq \sum_{x=1}^n \log (x) \leq \int_0^n \log  (x+1) \, dx
\\ 
\Leftrightarrow & n \log(n )-n+1 \leq \log n! \leq (n+1)\log(n+1)-n
\\ 
\Leftrightarrow & n \log(\frac ne )+1 \leq \log n! \leq (n+1)\log(\frac{n+1}{e})+1
\end{align}
Now apply the exponential function to get
\begin{align}
 & n \log(\frac ne )+1 \leq \log n! \leq (n+1)\log(\frac{n+1}{e})+1
\\ 
\Leftrightarrow &\Bigl(\frac ne\Bigr)^n e  \leq  n!\leq \Bigl(\frac{n+1}{e}\Bigr)^{n+1}e
\end{align}
This isn't exactly what you asked for, but consider the following:
\begin{align}
\Bigl(\frac{n+1}{e}\Bigr)^n =\Bigl(\frac{n}{e}\Bigr)^n\Bigl(\frac{n+1}{n}\Bigr)^n
=\Bigl(\frac{n}{e}\Bigr)^n\Bigl(1+\frac{1}{n}\Bigr)^n\leq\Bigl(\frac{n}{e}\Bigr)^n e
\end{align}
Because  $\Bigl(1+\frac{1}{n}\Bigr)^n$ is monotone increasing and converging towards $e$ (there are several proofs).
Thus
\begin{align}
\Bigl(\frac{n+1}{e}\Bigr)^n\leq \Bigl(\frac{n}{e}\Bigr)^n e \leq n!\leq \Bigl(\frac{n+1}{e}\Bigr)^{n+1}e
\end{align}
A: I was not able to complete the whole proof in detail, but here is a sketch anyway. There are some minor issues regarding what happens when $n =0,1$, but those two cases can be treated separately (and trivially). First a few observations


*

*$f(x)=\log_e(x)$ is an increasing function for $x>1$

*By considering rectangles of width 1 and height $f(x)$ from $x=1 $ to $n$, we see that
$$f(1)+f(2)+...+f(n-1)<\int_1^{n} \log_e(x)dx<f(1)+f(2)+...+f(n) $$


Using basic log laws for the outer terms, and evaluating the integral in the middle, we obtain
$$\log_e(n!)<\log_en^n-n+1<\log_e((n+1!))$$
Exponentiating and some rearrangements give
$$(n-1)!<\frac{n^{n}}{e^{n-1}}<n!$$
This gives us the inequality on the right $n!<\frac{(n+1)^{n+1}}{e^n}$ when we replace $n$ with $n+1$. We also have
$$\frac{n^{n}}{e^{n-1}}<n!$$
which is almost what we wanted, except that we need $\frac{(n+1)^n}{e^n}$ instead on the left hand side. It remains then to show that
$$\frac{(n+1)^n}{e^n}<\frac{n^{n}}{e^{n-1}}\Leftrightarrow (n+1)<(e^{\frac{1}{n}}n)$$
This is easily seen to be true if we look at the series expansion of  $(e^{\frac{1}{n}}n)-(n+1) $ (the first two terms of $(e^{\frac{1}{n}}n)$ cancels with $-(n+1)$)
