How to show $\lim_{k\rightarrow \infty} \left(1 + \frac{z}{k}\right)^k=e^z$ I need to show the following:
$$\lim_{k\rightarrow \infty} \left(1 + \frac{z}{k}\right)^k=e^z$$
For all complex numbers z. I don't know how to start this. Should I use l'Hopitals rule somehow?
 A: Presumably you know that $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)^n = e$. Try setting $n = \frac{k}{z}$ where $z$ is a constant and see where that gets you.
A: the following hints may suggest one line of approach. i will not spoil your fun by attempting to write a complete solution
if for positive integers $k \le m$ we set 
$$
f_{m,k}(z) = \binom{m}{k} \frac{z^k}{m^k}
$$
(note that $\frac{\binom{m}{k}}{m^k} \lt \frac1{k!}$)
then for each $k$, we have
$$
\lim_{m \to \infty} f_{m,k}(z) = \frac{z^k}{k!}
$$
with the convergence being uniform on any disk $|z| \le R$
now define for all $m \gt n$:
$$
F_{m,n}(z) = \sum_{k=0}^n f_{m,k}(z)
$$
so that for each $n$ we have
$$
F_n(z)=\lim_{m \to \infty} F_{m,n}(z) = \sum_{k=0}^n \frac{z^k}{k!}
$$ 
with the convergence again uniform on compact disks.
the binomial theorem gives:
$$
\left(1+\frac{z}{m} \right)^m = \sum_{k=0}^m \binom{m}{k} \frac{z^k}{m^k}
$$
so for $m \gt n$ we may write
$$
|(1+\frac{z}{m})^m - F_{m,n}(z)| \le \sum_{k=n+1}^m |f_{m,k}(z)| \lt \sum_{k=n+1}^m \frac{|z|^k}{k!}
$$
A: The binomial expansion of $\left(1+\frac{z}{k}\right)^k$ is
$$\begin{align}
\left(1+\frac{z}{k}\right)^k&=\sum_{j=0}^k{\binom{k}{j} \left(\frac{z}{k}\right)^j}\\\\
&=\sum_{j=0}^k \left( \frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)  \end{align}$$
The term $ \frac{k!}{(k-j)!\,k^j}$ can be expanded as
$$\begin{align}
\frac{k!}{(k-j)!k^j}&=\frac{k(k-1)\dotsb (k-j+1)}{k^j}\\\\
&=\left(\frac{k}{k}\right) \left(\frac{k-1}{k}\right)\left(\frac{k-2}{k}\right) \dotsb \left(\frac{k-(j-1)}{k}\right)\\\\
&=\left(1-\frac1k\right)\left(1-\frac2k\right) \dotsb \left(1-\frac{j-1}{k}\right)
\end{align}$$
and approaches $1$ as $k \rightarrow \infty$.  Thus,
$$\begin{align}
\lim_{k \to \infty} \left(1+\frac{z}{k}\right)^k&=\lim_{k \to \infty}\sum_{j=0}^k \left( \frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)\\\\
&=\lim_{k \to \infty}\sum_{j=0}^\infty \xi_{[0,k]}(j)\left(\frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)\\\\
& = \sum_{j=0}^\infty \frac{z^j}{j!}
\end{align}$$
which is the series representation for $e^z$!
Note that $\xi_{[0,k]}(j)$ is the indicator function and is $1$ when $j\in[0,k]$ and $0$ otherwise.  Also note that the last equality is justified since the series $\sum_{j=0}^\infty \xi_{[0,k]}(j)\left(\frac{k!}{(k-j)!\,k^j}\right) \left(\frac{z^j}{j!}\right)$ converges uniformly.
