I have a question about the following proofs how $f(z)$ and $e^{f(z)}$ can not have common poles:
Assume $z_0$ is a pole of $f$, then in a punctured disk near $z_0$ one can write $f(z) = \frac{g(z)}{(z-z_0)^{k}}$, g(z) holomorphic and $g(z_0) \neq 0$. Let $g(z_0) = re^{i \theta}$, and $z_n = z_0 + \frac{e^{i \theta/n}}{n}$. Then since $f(z_n)=g(z_n) e^{- i\theta} n^n $ and $\lim_{n \to \infty} g(z_n) = re^{i \theta}$, $\lim_{n \to \infty} e^{f(z_n)} = \infty$ So it follows that if $z_0$ a pole, it can not be a removable singularity of $e^f$. Further consider $$w_n = z_0 + \frac{e^{i (\theta+\pi)/n}}{n}$$ and we see that $\lim_{n \to \infty} e^{f(w_n)} = 0$
my question is that when calculating $f(z_n)$ why was the $z_n$ taken to a power of n? and where did $k$ the order of pole go? because I get $f(z_n) = \frac{g(z_n)}{({\frac{e^{i\theta/n}}{n})}^{k}}$
That reasoning is indeed wrong. A better choice is $$ z_n = z_0 + \frac 1n e^{i\theta/k} \, . $$ Then $$ f(z_n) = \frac{g(z_n) n^k}{e^{i \theta}} = \frac{g(z_n)}{g(z_0)} \cdot r n^k $$ and therefore $$ \left|e^{f(z_n)}\right| = e^{\operatorname{Re}f(z_n)} = e^{ rn^k \operatorname{Re}(g(z_n)/g(z_0))} > e^{r n^k/2} $$ for sufficiently large $n$. This shows that $\lim_{n\to \infty} e^{f(z_n)} = \infty$.
For the second part you would choose $$ w_n = z_0 + \frac 1n e^{i(\theta+ \pi)/k} \, . $$