To show $f(n)$ is monotonically increasing is equivalent to showing $f(n) < f(n+1)$ holds for all $n$. This can be done by applying the AM-GM inequality on the following $(n+1)$ positive numbers.

$$x_1 = 1,\quad x_2=x_3=\cdots=x_{n+1}=1+\frac{1}{n}$$

Because they are not all equal, inequality will strictly hold, i.e. $$\begin{align*} \sqrt[n+1]{x_1x_2x_3\cdots x_{n+1}} &\quad<\quad \frac{x_1+x_2+x_3+\cdots+x_{n+1}}{n+1}\\ \sqrt[n+1]{\left(1+\frac{1}{n}\right)^n} &\quad<\quad\frac{1}{n+1}\left(1 + n\left(1+\frac{1}{n}\right)\right)\\ \left(1+\frac{1}{n}\right)^{\frac{n}{n+1}} & \quad < \quad 1+\frac{1}{n+1}\\ \left(1+\frac{1}{n}\right)^n & \quad < \quad\left(1+\frac{1}{n+1}\right)^{n+1} \end{align*}$$

Next, we are going to show that 

$$\left(1+\frac{1}{n}\right)^n < 3$$

Applying binomial expansion gives

$$\begin{align*} \left(1+\frac{1}{n}\right)^n &=\ 1+\binom{n}{1}\frac{1}{n}+\sum_{k=2}^{n}\binom{n}{k}\frac{1}{n^k}\\ &=\ 1 + 1 + \sum_{k=2}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\cdot\frac{1}{n^k}\\ &=\ 2 + \sum_{k=2}^{n}\frac{1}{k!}\cdot\frac{n(n-1)\cdots(n-k+1)}{n^k}\\ &\le\ 2 + \sum_{k=2}^{n}\frac{1}{k!}\\ &\le\ 2 + \sum_{k=2}^{n}\frac{1}{k(k-1)}\\ &=\ 2 + \sum_{k=2}^{n}\left(\frac{1}{k-1}-\frac{1}{k}\right)\\ &=\ 2 + \left(\frac{1}{2-1} - \frac{1}{n}\right) <\ 3\end{align*}$$

Therefore, we conclude that $\displaystyle\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n$ must exist.