Why is $\exp(z^{-1})$ an essential singularity?

I'm reading Conway's complex analysis book and on page 105 he defines a pole as:

Definition: If $z=a$ is an isolated singularity of $f$, then $a$ is a pole of $f$ if $\lim_{z\to a}|f(z)|=\infty$.

There is also a characterization theorem on page 103:

Theorem: If $f$ has an isolated singularity at $a$ then the point $z=a$ is a removable singularity iff $$\lim_{z\to a}(z-a)f(z)=0$$

I'm trying to use these facts to prove the assertion he made on page 105:

Knowing that an essential singularity is neither a pole nor a removable singularity, how can I prove $\exp(z^{-1})$ has an essential singularity at $z=0$?

• I start with essential singularity = isolated singularity which is not dominated by any power of $(z-a)$, i.e. there is a path to $a$ where $|f(z)|$ grows much faster than $(z-a)^k$ for any $k$. Then, I prove that every isolated singularity is a pole or an essential singularity (a removable singularity being some sort of pole of order $0$). All this reduce to proving the Riemann theorem on removable singularities , the Laurent series theorems, and the Liouville theorem. – reuns May 14 '16 at 20:19
• Use its Laurent expansion. – Starfall May 14 '16 at 20:21
• $z=0$ is not a pole, since the limit from the left is $0$ and not $\infty$. It is not a removable singularity since the limit from the right of $zf(z)$ is not $0$. – GEdgar May 14 '16 at 20:23
• @Starfall : If you already proved everything, it is obvious that it is an essential singularity. But what do you prove before saying "use its Laurent expansion" ? – reuns May 14 '16 at 20:24
• The definition of $\exp$ is usually gvien by the Taylor series, so it should be trivial to note that the Laurent expansion has a singularity at $z = 0$ which is not of finite order. – Starfall May 14 '16 at 20:28