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 $\frac{n(n-1)\ldots\big(n-(k-1)\big)}{n^k}<1$,
$$\begin{align*}
a_n&<1+\left(1+\frac12+\frac1{2^2}+\ldots+\frac1{2^{n-1}}\right)\\
&<1+\left(\frac{1-\left(\frac12\right)^n}{1-\frac12}\right)<3-\frac1{2^{n-1}}<3
\end{align*}$$

Hint Let
$$U_n = (1 + 1/n)^n \\
V_n=U_n = (1 + 1/n)^{n+1}$$
It is clear that $U_n < V_n$ for all $n$. Use Bernoulli inequality to show that
$$\frac{U_{n+1}}{U_n} \geq 1 \\
\frac{V_{n}}{V_{n+1}} \geq 1$$
Find some $m$ so that $V_m \leq 3$.
Deduce from here that
$$U_1 \leq U_n \leq V_m$$ for all $n$.
