Show that $2\le U_n \le 3$ for all $n$ Given $U_n = (1 + 1/n)^n$ , $n = 1,2,3,\ldots\;.$

Show that $2 \le U_n \le 3$ for all n

This is what I've done. Can anyone help?
$$\begin{align*}
a_n=\left(1+\frac1n\right)^n&=\sum_{r=0}^n{^nC_r}(1)^r\left(\frac1n\right)^{n-r}\\
&=\sum_{r=0}^n{^nC_r}\left(\frac1n\right)^{n-r}\\
&=1+\frac{n}n+\frac1{2!}\frac{n(n-1)}{n^2}+\frac1{3!}\frac{n(n-1)(n-2)}{n^3}\\
&\quad\,+\ldots+\frac1{n!}\frac{n(n-1)\ldots(n-n+1)}{n^n}
\end{align*}$$
Since $\forall k\in\{2,3,\ldots,n\}$: $\frac1{k!}<\frac1{2^k}$, and $\frac{n(n-1)\ldots\big(n-(k-1)\big)}{n^k}<1$,
$$\begin{align*}
a_n&<1+\left(1+\frac12+\frac1{2^2}+\ldots+\frac1{2^{n-1}}\right)\\
&<1+\left(\frac{1-\left(\frac12\right)^n}{1-\frac12}\right)<3-\frac1{2^{n-1}}<3
\end{align*}$$

Here is another way to look at it that might be helpful, if the question is why for all $n$ it holds. Consider $f(x)=(1+\frac{1}{x})^x$. It is an increasing function of $x$:
$$ f'(x)=\frac{d}{dx}e^{x \ln(1+\frac{1}{x})} 
= \left[\ln\left(1+\frac{1}{x}\right) - \frac{1}{1+x}\right]\left(1+\frac{1}{x}\right)^x >0, \forall x>0$$
Now:
$f(1)=2$, and $\lim_\limits{n\rightarrow+\infty}f(n)=e<3$, so $2\leq f(n) < e < 3, \forall n$
