 An exponential has no isolated pole

If $g$ is holomorphic on a punctured disc about $a$, the function $e^g$ cannot have a pole at $a$. Indeed, $(e^g{)}^{\prime }/e^g=g^{\prime }$, whereas a pole of order $n\ge 1$ would give residue $-n$ for this logarithmic derivative. The Laurent series derivative of $g$ has residue zero, since its only candidate term comes from differentiating the constant coefficient. This contradiction rules out a pole, regardless of the type of the isolated singularity of $g$.