Use mathematical induction to prove that for any $k \in\mathbb N , \lim (1+k/n)^n = e^k$. Use mathematical induction to prove that for any $k \in \mathbb N, \lim (1+k/n)^n = e^k$.
I already used monotone Convergence Theorem to prove $k=1$ case. Do I just need to go through the same process to show $k$? If not, could you please help?
Thanks

First a useful bound: Suppose $\theta_n \ge 0$ is such that $\lim_n n \theta_n = 0$, then $\lim_n (1+ \theta_n)^n = 1$. To see this, note $(1+\theta_n)^n = \sum_{k=0}^n \binom{n}{k} \theta^k \le \sum_{k=0}^n n^k \theta^k  \le \frac{1}{1-n \theta_n}$.
Suppose the result is true for $k$ (I am taking $k=1$ as already proved). Then note that 
$(1+ \frac{k}{n}) (1+ \frac{1}{n}) = (1+ \frac{k+1}{n}+ \frac{k}{n^2}) = (1+ \frac{k+1}{n})(1+ \frac{k}{n^2 (1+ \frac{k+1}{n})}) $, and let $\theta_n = \frac{k}{n^2 (1+ \frac{k+1}{n})}$. This gives
$  (1+ \frac{k+1}{n})^n = \frac{1}{(1+\theta_n)^n} (1+ \frac{k}{n})^n (1+ \frac{1}{n})^n$. Taking limits gives
$ \lim_n (1+ \frac{k+1}{n})^n  = \frac{1}{1}e^k e^1  = e^{k+1}$, hence the result is true for $k+1$.
This was my original answer which is not correct for the reason Robjohn points out below:
Assuming that you have proved it for $k=1$, note that the function $x \mapsto x^k$ is continuous, and
$(1+ \frac{k}{n})^n = (1+ \frac{k}{n})^{\frac{n}{k}k} = \left ( (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k$.
Hence $\lim_n (1+ \frac{k}{n})^n = \lim_n \left ( (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k = \left( \lim_n (1+ \frac{k}{n})^{\frac{n}{k}} \right)^k = \left( \lim_n (1+ \frac{1}{\frac{n}{k}} )^{\frac{n}{k}} \right)^k  = e^k $.
