Show that $0$ is an essential singularity of $e^{\frac{1}{z}}$

Question

I am working on the following exercise:

Show that $0$ is an essential singularity of $e^{\frac{1}{z}}$.

I am new to Laurent Series, so I would be grateful if you could have a look at my attempt:

I would say that it is enough to note that (by definition) $$e^{\frac{1}{z}} = \sum_{n=0}^\infty \frac{1}{z^n n!},$$ which is already the Laurent Series of $e^{\frac{1}{z}}$. It is obvious that the main part of this Laurent Series does not vanish, so the singularity is essential. Is this enough or did I forget to check any preconditions?

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, it is enough what you have done.