Prove that $a_n=(1-\frac{1}{n})^n$ is monotonically increasing sequence I try to solve it Bernoulli inequality but it too complicated, am I missing something easier?
My try-
$$\frac{a_n}{a_{n+1}}=\frac{(1-\frac{1}{n})^n}{(1-\frac{1}{n+1})^{n+1}}\\=(\frac{1}{1-\frac{1}{n+1}})(\frac{\frac{n-1}{n}}{\frac{n}{n+1}})^n\\=(\frac{1}{1-\frac{1}{n+1}})(1-\frac{1}{n^2})^n<1\\\iff (1-\frac{1}{n^2})^n<1-\frac{1}{n+1}$$
here is where that I want to use Bernoulli...

$a_n:= (1-1/n)$, $n\in \mathbb{Z+}.$
$a_{n+1} = (1-1/(n+1)).$
$a_n \lt a_{n+1}.$
$\rightarrow:$
$(a_n)^n < (a_{n+1})^n.$
Since $1> a_n  >0 :$
$a_n (a_n)^n < (a_{n+1})^n, $
$(a_n)^{n+1} < (a_{n+1})^n.$
Set $s:= \dfrac{1}{n(n+1)} >0$,  
Note : $ 0 <s<1$.
Consider $f(x) := x^s,$  $x >0$, real.
$f'(x) = s\dfrac{1}{x^{1-s}}>0.$
Hence $f(x)$ is strictly increasing.
$\Rightarrow:$
$f((a_n)^{n+1}) = (a_n)^{1/n} < $
$f((a_{n+1})^n = (a_{n+1})^{1/(n+1)},$
strictly increasing.
