equality of complex analysis property I need to prove the following that
$e^{\bar{z}} = \bar{e^{z}}$ 
with $e^z :=  \Sigma_{k = 0}^{k = \infty} \frac{z^k}{k!}$
The problem that I am having is first we know $e^z$ defined that way always converge, but what I don't understand is we have to see that $\bar{z^k}$ put into the formula above converges to same number as if we do the summation and then converge to a number, and conjugate that. I don't know how to do that..

We know that 
$$\sum_{k=0}^\infty\frac{z^k}{k!}$$
converges for all $z\in\Bbb C$.  Let's let $a_k=1/k!$, and $f(z)=e^z$.
Now,
$$\begin{align}\overline{f(z)} & =\overline{\sum_{k=0}^\infty a_kz^k}\\
& = \sum_{k=0}^\infty a_k\overline z^k,\quad\text{because }a_k\in\Bbb R\\
& = f(\overline z).
\end{align}$$
