How is $a_n=(1+1/n)^n$ monotonically increasing and bounded by $3$?

I was reading about how completeness is required for limits. And I came across this:

the sequence $a_n=(1+1/n)^n$ is monotonically increasing and bounded by 3 and so we expect it to converge, but that it does not converge within $\mathbb{Q}$. More generally it stands to reason that any sequence of real numbers which is increasing and bounded must converge to some real number. This is a consequence of completeness of $IR$

My question is: How is the mentioned sequence monotonically increasing and bounded by $3$ ?

The monotonicity follows from the AM-GM inequality for $n+1$ points. Taking $x_1 = x$ and $x_2 = \cdots = x_{n+1} = y$, we get $$ \sqrt[n+1]{xy^n} \leq \frac{x+ny}{n+1}. $$ In particular, taking $x = 1$ and $y = 1+\frac{1}{n}$ yields $a_n \leq a_{n+1}$ ($n \geq 1$).

Now as another special case, take $x = 1$ and $y = 1-\frac{1}{n}$ to get $$ b_n \geq b_{n+1} \qquad (n \geq 1), $$ where $b_n := \left( 1-\frac{1}{n}\right)^{-n}$. We let you verify that $$ b_n = a_{n-1} \left( 1+\frac{1}{n-1}\right) \geq a_{n-1}, \qquad (n \geq 2). $$ Hence $$ a_1 \leq a_2 \leq a_3 \leq \cdots \leq b_3 \leq b_2 \leq b_1 $$ and we deduce that $a_n \leq b_m$ for all $n, m \in \mathbb{N}$. In particular, taking $m = 6$, we get $$ a_n \leq \left( \frac{6}{5} \right)^6 \leq 3. $$