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.

A tricky way to verify the monotonicity is by applying AM-GM inequality as follows:
\begin{align*}
a_n & = \left(1 + \frac{1}{n}\right)^n \\
& = 1\times \left(1 + \frac{1}{n}\right) \times \cdots \times \left(1 + \frac{1}{n}\right)\\
& \leq \left[\frac{1 + n\left(1 + \frac{1}{n}\right)}{n + 1}\right]^{n + 1} \\
& = \left(1 + \frac{1}{n + 1}\right)^{n + 1} = a_{n + 1}.
\end{align*}
