This question was part of an assignment that couldn't be discussed due to the pandemic.
Question: Let $f$ have an isolated singularity at a point $a$. Prove that $e^{f}$ cannot have a pole at point $a$.
Attempt: Let on the contrary it has a pole of the order say $k.$ Then $e^f {(z-a)^k}$ is analytic at a and $\lim_{z\to a} f(z) = M$.
But I am unable to see a contradiction based on this.
Can someone please tell which result should I use to obtain a contradiction?
Thank you!!