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: You brought the wrong thing to the right hand side. Your inequality is equivalent to
$$1 + \frac{1}{n} < \biggl(1 + \frac{1}{n^2-1}\biggr)^n\,.$$
Now applying Bernoulli's inequality to the right hand side we get
$$\biggl(1 + \frac{1}{n^2-1}\biggr)^n \geqslant 1 + \frac{n}{n^2-1} > 1 + \frac{1}{n}\,.$$
A: With Bernoulli, but rewriting to make it simpler:
\begin{align}
&\Bigl(1-\frac1{n+1}\Bigr)^{n+1}>\Bigl(1-\frac1{n}\Bigr)^{n}\iff\Bigl(\frac n{n+1}\Bigr)^{n+1}>\Bigl(\frac{n-1}{n}\Bigr)^{n}\\
\iff&\Bigl(\frac{n^2}{n^2-1}\Bigr)^{n}>\frac{n+1}n=1+\frac1n\iff\Bigl(1+\frac{1}{n^2-1}\Bigr)^{n}>1+\frac1n.
\end{align}
Now, by Bernoulli's inequality,
$$\Bigl(1+\frac{1}{n^2-1}\Bigr)^{n}>1\frac{n}{n^2-1}>1+\frac{n}{n^2}=1+\frac1n.$$
A: Let $y=x\ln\left(1-\dfrac1x\right)$ so $y'>0$ in $(1,\infty)$ and with mean-value theorm on $[n,n+1]$ we see
$$f(n+1)-f(n)=f'(\xi)~~~\text{for a}~\xi\in(n,n+1)$$
with substitution 
$$(n+1)\ln\left(1-\dfrac{1}{n+1}\right)-n\ln\left(1-\dfrac1n\right)>0$$
and we find the result!
A: $a_n:= (1-1/n)$, $n\in \mathbb{Z+}.$
$a_{n+1} = (1-1/(n+1)).$
$a_n \lt a_{n+1}.$
$\rightarrow:$
$(a_n)^n < (a_{n+1})^n.$
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.
