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?

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
$$
|(1+\frac{z}{m})^m - F_{m,n}(z)| \le \sum_{k=n+1}^m |f_{m,k}(z)| \lt \sum_{k=n+1}^m \frac{|z|^k}{k!}
$$
