Show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing. 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.

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 $$1 + \frac{n}{{\left( {n + 1} \right)\left( {n - 1} \right)}} - \left( {1 + \frac{1}{n}} \right) = \frac{1}{{n\left( {n - 1} \right)\left( {n + 1} \right)}} > 0\;;\; \text{if }\;n\geqslant 2$$

$(1)$ $$\begin{align}
   {\left( {1 - \frac{1}{n}} \right)^{ - n}} &> {\left( {1 - \frac{1}{{n + 1}}} \right)^{ - n - 1}}  \\
   {\left( {\frac{{n - 1}}{n}} \right)^{ - n}} &> {\left( {\frac{n}{{n + 1}}} \right)^{ - n}}{\left( {\frac{n}{{n + 1}}} \right)^{ - 1}}  \\ 
   {\left( {\frac{{n - 1}}{n}} \right)^{ - n}} &> {\left( {\frac{n}{{n + 1}}} \right)^{ - n}}\left( {1 + \frac{1}{n}} \right)  \\
   {\left( {\frac{n}{{n - 1}}} \right)^n}{\left( {\frac{n}{{n + 1}}} \right)^n} &> 1 + \frac{1}{n}  \\
   {\left( {\frac{{{n^2}}}{{{n^2} - 1}}} \right)^n} &> 1 + \frac{1}{n} \\
   {\left( {1 + \frac{1}{{{n^2} - 1}}} \right)^n} &> 1 + \frac{1}{n} \end{align} $$
