Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\frac{1}{(n+1)^2} \right )^n\left ( \frac{n+2}{n+1} \right )$$
Using the Bernoulli inequality, we see that, for all $n\geq 1$,
$$\left ( 1-\frac{1}{(n+1)^2} \right )^n\geq 1-\frac{n}{(n+1)^2}=\frac{n^2+n+1}{n^2+2n+1}.$$
How do you then show that 
$$\left ( 1-\frac{n}{(n+1)^2} \right )\left ( \frac{n+2}{n+1} \right )>1?$$
Edit: This is not a duplicate question; the question is not to show about the existence of the definition of Euler number $e$. The question is about showing that it is increasing, the way I have shown that I have been stuck with, not other ways. It seems that the question is too easy that I have been too tired to think at this late.

It can be proved using function. We have $f(x) = \left(1+\dfrac{1}{x}\right)^x, x \in [1,\infty)$ and taking log: $\log f(x) = x\log(x+1) - x\log x\implies f'(x) = f(x) \left(\log(x+1)+ \dfrac{x}{x+1} - \log x - 1\right)= f(x)\left(\log(x+1) - \log x - \dfrac{1}{x+1}\right)= f(x)g(x)\implies g'(x) = \dfrac{1}{x+1}-\dfrac{1}{x}+\dfrac{1}{(1+x)^2}= \dfrac{x(x+1)-(x+1)^2+x}{x(1+x)^2}= -\dfrac{1}{x(1+x)^2} < 0\implies g(x) > \displaystyle \lim_{x \to \infty} g(x)=0\implies f'(x) > 0\implies f(n+1) > f(n)$.
