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.

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
$$\frac{a_n}{a_{n+1}}  \geq \left( 1+\frac{n+1}{n^2-1} \right) \frac{n+1}{n}=\frac{n}{n-1}  \frac{n-1}{n}=1$$
