Show that if $f(z)$ is a non-constant entire function , then $g(z)=exp(f(z))$ has essential singularity in $z=\infty$.
This question has been answered in the link Thanks!
My approach: Let $f$ analytic (then $f$ has a series), then $exp(f(z))$ is analytic. Then
$$exp(f(\frac{1}{w}))=\sum_{n=0}^{\infty}{\frac{1}{n!}(f(\frac{1}{w}))^{n}}=\sum_{n=0}^{\infty}{\frac{1}{n!}(\sum_{m=0}^{\infty}{\frac{a_{m}}{w^m}})^{n}}$$
We can see that $exp(f(\frac{1}{w}))$ has essential singularity in $w=0$, since there exist negative coefficients. So, $exp(f(z))$ has essential singularity in $w=0$. Is this right? Thanks!
If the singularity at infinity were removable, then $f$ would be bounded. Then by Liouville's theorem is constant. See this answer.
So it appears we can get that it's a non-removable singularity. But as @N.S. points out, it could still be a pole. To rule that out, see this.