Prove $(1 + \frac{1}{n})^n$ is bounded above

I've checked similar questions on the site but couldn't find satisfactory solutions or hints.

Also, is there a more general approach to proving whether a given sequence is bounded below or above?

By the binomial formula: \begin{eqnarray*} \left(1+\frac{1}{n}\right)^n=1+1+\sum_{k=2}^n\binom{n}{k}\cdot\left(\frac{1}{n}\right)^k. \end{eqnarray*}

Notice that \begin{eqnarray*} \binom{n}{k}\cdot\left(\frac{1}{n}\right)^k=\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac{1}{k!}<\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}. \end{eqnarray*}

So we get \begin{eqnarray*} \left(1+\frac{1}{n}\right)^n<2+\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=3-\frac{1}{n}<3. \end{eqnarray*}

Hint: Use the binomial formula:

$$\left(1 + \frac{1}{n}\right)^n =\sum_{j=0}^n\frac1{j!}\frac{n!}{(n-j)!}\frac1{n^j}= 1+1+\sum_{j=2}^n\frac1{j!} \prod_{k=1}^{j-1}(1-k/n) < \ldots$$