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..

First, we have
$$\overline{\lim_{N\to \infty}S_N}=\lim_{N\to \infty}\overline{S_N}$$
Second, we have
$$\overline{S_N}=\overline{\sum_{k=0}^Nf_k(z)}=\sum_{k=0}^N\overline{f_k(z)}$$
Third, we have
$$\overline{f_k(z)}=\overline{\left(\frac{z^k}{k!}\right)}=\frac{\overline{z^k}}{k!}$$
Finally, we have
$$\overline{z^k}=\bar z^k$$
and we are done, as this last equality follows inductively from the fact that $\overline{z_1z_2}=\bar z_1\bar z_2$!  
