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Show that if $f(z)$ is a non-constant entire function, then $e^{f(z)}$ has an essential singularity at $z=\infty$.

This is my approach:
By Liouville's theorem I know that if $f$ is a non-constant entire function it is not bounded, then it has a (no removable) singularity at $z=\infty$.
I can show that if $f$ has an essential singularity then $e^{f(z)}$ has an essential singularity, by using Casorati-Weierstrass theorem. The rest is show that if $f$ has a pole then $e^{f}$ has an essential singularity; actually I don't have idea to do this. Any hint?

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  • $\begingroup$ I think I can use the fact that $f(z)$ has a pole at $z_{0}$ iff the function $\frac{1}{f(z)}$ has a zero at $z_{0}$ and use the power series of exponential function at infinity, but is just an idea. $\endgroup$ – Alberto Navarro May 13 '18 at 19:00
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It suffices to show that $h(z)=e^{f(1/z)}$ has an essential singularity at $0$. Obviously $h$ has an isolated singularity at $0$.

Case 1: this singularity is removable. Then $h(z)$ is bounded around $0$. So there exist $M,r>0$ such that when $0<|z|<r$ it is $|h(z)|<M$. Setting $w=1/z$, we get that $|e^{f(w)}|< M$ for $|w|>1/r$. The function $z\mapsto e^{f(z)}$ is entire and therefore continuous, so it is bounded on the compact set $\overline{D(0,1/r)}$. But we just saw that it is bounded on the complement of this disk too, so it is bounded on $\mathbb{C}$, which is impossible by Liouville's theorem.

Case 2: This singularity is a pole of order $m$. Then $\displaystyle{\lim_{z\to0}h(z)=\infty}$, that is $\displaystyle{\lim_{z\to0}e^{f(1/z)}}=\infty$. But now we can consider $g(z)=e^{-f(1/z)}$ and it is $\displaystyle{\lim_{z\to 0}g(z)=0}$, hence $g$ has a removable singularity at $0$. Working as in case 1, we get the desired contradiction.

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