Limit with number $e$ and complex number This is my first question here. I hope that I spend here a lot of fantastic time.
How to proof that fact?
$$\lim_{n\to\infty} \left(1+\frac{z}{n}\right)^{n}=e^{z}$$
where $z \in \mathbb{C}$ and $e^z$ is defined by its power series.
I have one hint: find the limit of abs. value and arguments,
but i don't know how to use it to solve that problem.
Thank you for help.
Before I try solve this problem, I proofed that $$e^z=e^{x}(\cos y + i \sin y)$$
where $z=x+yi$ ,maybe this help.

Here is an approach.

$$\lim_{n\to \infty} \left(1+\frac{z}{n}\right)^{n} =\lim_{n\to \infty} \sum_{k=0 }^{n} {n\choose k} \frac{z^k}{n^k}. $$

Now try to prove that 

$$\lim_{n\to \infty} \frac{{n\choose k}}{n^k} = \frac{1}{k!}. $$

Note:

$$ {n\choose k} = \frac{n!}{(n-k)! k!}. $$

