It is a well known fact that $0$ is an essential singularity of the function $e^{1/z}$ therefore it is neither removable nor a pole.
On the other hand we know the Rieamann's continuation theorem which states that an holomorphic function $f:U\setminus \{z_0\}\to \mathbb{C}$ has a removable singularity at $ z_0$ if and only if $f(z) $ is bounded in some neighborhood of $ z_0$.
I'm confussed because of the following calculation:
If we pick $\varepsilon>0 $ and $ z$ such that $|z|<\varepsilon $ then
$$ |z e^{1/z} |< \varepsilon \sum^\infty_{k=0} \left|\frac{1}{k!z^k} \right|=\varepsilon \sum^\infty_{k=0} \frac{1}{k!\varepsilon^k} =\varepsilon e^{1/\varepsilon}$$
The previous means that $ 0$ is a removable singularity of $ ze^{1/z}$ which implies that $ 0$ is a pole of order 1 of $ e^{1/z}$ which cannot be since the well known fact of the begining.
Of course I'm wrong in something but I couldn't tell in what.
Thank you all in advance
A (possibly) useful definition
My definition of a pole is:
$z_0 $ is a pole of order $ m$ of the holomorphic function $f:U\setminus \{z_0\}\to \mathbb{C}$ if $m $ is the minimum integer such that $(z-z_0)^mf(z) $ has a removable singularity at $ z_0$.
Best Answer
The first equality in the displayed equation is wrong. You replaced $|z|$ by $\epsilon$, but $|z|$ can be arbitrarily small, and hence the summands can be arbitrarily large.