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
 A: 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 $.
A: Suppose
$$
\lim_{n\to\infty}\left(1+\frac1n\right)^n=e
$$
Bernoulli's Inequality (which is proven by induction at the end of this answer) yields
$$
\begin{align}
\frac{\left(\color{#C00000}{1+\frac k{n+1}}\right)^{n+1}}{\left(\color{#00A000}{1+\frac kn}\right)^n}
&=\color{#00A000}{\frac{n+k}{n}}\left(\color{#C00000}{\frac{n+k+1}{n+1}}\color{#00A000}{\frac{n}{n+k}}\right)^{n+1}\\
&=\frac{n+k}{n}\left(1-\frac{k}{(n+1)(n+k)}\right)^{n+1}\\
&\ge\frac{n}{n+k}\left(1-\frac{k}{n+k}\right)\\
&=1
\end{align}
$$
That is, $\left(1+\frac kn\right)^{n/k}$ is increasing in $n$.
Since $\left(1+\frac 1n\right)^{n}$ is a subsequence of the increasing sequence $\left(1+\frac kn\right)^{n/k}$, they both tend to the same limit. That is,
$$
\lim_{n\to\infty}\left(1+\frac kn\right)^{n/k}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e
$$
which implies
$$
\lim_{n\to\infty}\left(1+\frac kn\right)^n=e^k
$$
A: A similar yet different approach:
Assume that $\lim_{n\to\infty}\left(1+\frac kn\right)^n=e^k$ for some $k$ and prove it for $k+1$. Notice that
$$\begin{align}
1+\frac{k+1}{n} &= \frac{n+k+1}{n}
\\ &=\frac{n+k+1}{n+k}\cdot \frac{n+k}{n}
\\ &=\left(1+\frac{1}{n+k}\right)\cdot\left(1+ \frac{k}{n}\right)
\end{align}$$
Now,
$$\begin{align}
\left(1+\frac{1}{n+k}\right)^n
&=\frac{\left(1+\frac{1}{n+k}\right)^{n+k}}{\left( 1+\frac{1}{n+k}\right)^k} 
\end{align}$$
The numerator is a subsequence of $(1+\frac{1}{n})^n$ and therefore converges to $e$, and the denominator converges to $1$. Then the fraction tends to $e$ when $n\to\infty$. From limit arithmetic, we have
$$\begin{align}
\lim_{n\to\infty}\left(1+\frac{k+1}{n}\right)^n 
&=\lim_{n\to\infty}\left(1+\frac{1}{n+k}\right)^n\cdot\left(1+ \frac{k}{n}\right)^n \\
&=e\cdot e^{k} \\
&=e^{k+1}.
\end{align}$$
