How do you show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing?
I understand that the binomial theorem should be used here but I don't see how we can use it to prove that $a_{n+1} < a_n$.
I will rewrite the sequence as, \begin{align*} (1 - \frac{1}{n})^{-n} &= (\frac{n-1}{n})^{-n} \\ &= (\frac{n}{n-1})^n \\ &= (1 + \frac{1}{n-1})^n \end{align*}
Then I can apply binomial theorem to it.
This is as far as I got now.
Let
$$a_n=(1 - \frac{1}{n})^{-n}=\frac{n^n}{(n-1)^n}$$
Then
$$\frac{a_n}{a_{n+1}}=\frac{n^n}{(n-1)^n}\frac{n^{n+1}}{(n+1)^{n+1}}=\frac{n^{2n+2}}{(n-1)^{n+1}(n+1)^{n+1}}\frac{n-1}{n}$$ $$=\left( \frac{n^{2}}{n^2-1} \right)^{n+1} \frac{n-1}{n}=\left( 1+\frac{1}{n^2-1} \right)^{n+1} \frac{n-1}{n}$$
By Bernoully or Binomial Theorem
$$\frac{a_n}{a_{n+1}} \geq \left( 1+\frac{n+1}{n^2-1} \right) \frac{n+1}{n}=\frac{n}{n-1} \frac{n-1}{n}=1$$
Ultimately, you want to show that$^{(1)}$ for $n\geqslant 2$ $${\left( {1 + \frac{1}{{\left( {n - 1} \right)\left( {n + 1} \right)}}} \right)^n} > 1 + \frac{1}{n}$$
Using the Binomial Theorem, the left hand side is $${\left( {1 + \frac{1}{{\left( {n - 1} \right)\left( {n + 1} \right)}}} \right)^n} > 1 + \frac{n}{{\left( {n - 1} \right)\left( {n + 1} \right)}}$$
Can you show the right hand side is $>1+n^{-1}$? Note that $$1 + \frac{n}{{\left( {n + 1} \right)\left( {n - 1} \right)}} - \left( {1 + \frac{1}{n}} \right) = \frac{1}{{n\left( {n - 1} \right)\left( {n + 1} \right)}} > 0\;;\; \text{if }\;n\geqslant 2$$
$(1)$ $$\begin{align} {\left( {1 - \frac{1}{n}} \right)^{ - n}} &> {\left( {1 - \frac{1}{{n + 1}}} \right)^{ - n - 1}} \\ {\left( {\frac{{n - 1}}{n}} \right)^{ - n}} &> {\left( {\frac{n}{{n + 1}}} \right)^{ - n}}{\left( {\frac{n}{{n + 1}}} \right)^{ - 1}} \\ {\left( {\frac{{n - 1}}{n}} \right)^{ - n}} &> {\left( {\frac{n}{{n + 1}}} \right)^{ - n}}\left( {1 + \frac{1}{n}} \right) \\ {\left( {\frac{n}{{n - 1}}} \right)^n}{\left( {\frac{n}{{n + 1}}} \right)^n} &> 1 + \frac{1}{n} \\ {\left( {\frac{{{n^2}}}{{{n^2} - 1}}} \right)^n} &> 1 + \frac{1}{n} \\ {\left( {1 + \frac{1}{{{n^2} - 1}}} \right)^n} &> 1 + \frac{1}{n} \end{align} $$
$$ \begin{align} \left(1-\frac1n\right)^{-n} &=\left(\frac{n}{n-1}\right)^n\\ &=\left(1+\frac1{n-1}\right)^n\\ \end{align} $$ and this was shown to be decreasing in this answer.
$\displaystyle a_n=\left(\frac{n}{n-1}\right)^n$
$\displaystyle \frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}(n-1)^n}{n^{2n+1}}<1$
Analytical method. Set $$ f(x)=\left(1-\frac{1}{x}\right)^{-x} =\exp\left(-x\log\left(1-\frac{1}{x}\right)\right). $$ Then $$ f'(x)=f(x)\left(\log\left(\frac{x}{x-1}\right)-\frac{1}{x-1}\right) $$ and one has to look where the derivative is negative, that is $$ \log\frac{x}{x-1}-\frac{1}{x-1}<0 $$ or, setting $t=x/(x-1)$, $$ \log t-(t-1)<0, $$ which is the case for every $t>0$ except $t=1$. Thus the original function is decreasing in every interval it is defined, in particular in the interval $(1,\infty)$.
Taking the $\log$ of the $n$th term of the sequence we get $$ b_n := -n \log \left(1 - \frac1n\right) = n \log n - n \log (n - 1) $$
We will show $b_n$ is decreasing, implying $a_n$ is decreasing as well. The derivative of $b_n$ (treating it as a real function) is \begin{align*} 1 + \log n - \log(n-1) - \frac{n}{n-1} &= \log n - \log(n-1) - \frac{1}{n-1} \\ &= - \int_{n-1}^{n} \left( \frac{1}{n-1} - \frac{1}{x}\right) \; dx \end{align*} which is less than $0$ for all $n \in \mathbb{R}$ since the integrand is positive. Since the derivative of $b_n$ is always negative, $b_n$ is decreasing and we are done.
