Proof $a_{n+1} \geq a_n$ with $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$ Let $a_n := (1+1/n)^n$ for $n \in \mathbb{N}$
How can one prove that $a_{n+1} \geq a_n$ for all $n \in \mathbb{N}$ with Bernoulli's inequality?
I know that the inequality states that $(1+x)^r \geq 1+rx$ for every integer $r \geq 0$ and every real number $x \geq -2$. And if the exponent $r$ is even, then the inequality is valid for all real numbers x. 
So I have to use induction and for $n=1$ we would get $(1+1/1)^1 = (1+1/(1+1))^2$, and that would give $2 < 2,25$. But wouldn't that imply that all numbers $> 1$ would make $a_{n+1} > a_n$? At which case is it equal? Can someone show me where I can find an induction proof for this?

$$
a_{n+1}\ge a_n \quad\Longleftrightarrow\quad 
\frac{(n+2)^{n+1}}{(n+1)^{n+1}}\ge\frac{(n+1)^n}{n^n}
\quad\Longleftrightarrow\quad \frac{n+2}{n+1}\ge\left(\frac{n^2+2n+1}{n^2+2n}\right)^n
\\ \quad\Longleftrightarrow\quad \left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{1}{n+2}
$$
The last inequality holds since
$$
\left(1-\frac{1}{n^2+2n+1}\right)^n\ge 1-\frac{n}{n^2+2n+1}
$$
and 
$$
\frac{1}{n+2}\ge\frac{n}{n^2+2n+1}
$$
