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.

Expand using the binomial formula: $\displaystyle \left(1+\frac{z}{n}\right)^n = \sum_{k=0}^n {n\choose k}\left( \frac{z}{n}\right)^k = \sum_{k=0}^\infty E_k^n$ where we define $\displaystyle E_k^n = {n\choose k}\left( \frac{z}{n}\right)^k$ for $k \le n$ and $= 0$ otherwise
We want $\displaystyle \sum_{k=0}^\infty E_k^n$ to converge to $\displaystyle \sum _{k=0}^\infty \frac{z^k}{k!}$ as $n \to \infty$ To do that we will show $\displaystyle E^n_k \to \frac{z^k}{k!}$ as $n \to \infty$
$\displaystyle E^n _k = \frac{n!}{k!(n-k)!}\left( \frac{z}{n}\right)^k = \frac{n!}{k!(n-k)!}  \frac{z^k}{n^k} = \frac{n!}{n^k (n-k)!}\frac{z^k}{k!} =\frac{n}{n} \frac{(n-1)}{n}\cdot \ldots \cdot \frac{(n-k+1)}{n}\frac{z^k}{k!} $
Therefore we just have to prove $\displaystyle \frac{n}{n} \frac{(n-1)}{n}\cdot \ldots \cdot \frac{(n-k+1)}{n} \to 1$. The number of terms to multiply is constant and equal to $k$. So there is no problem with invoking how each of them goes to $1$ seprately, and that limits commute with multiplication. 
