Laurent series for $e^\frac{1}{z}$ about $z=0$

Question

What I've currently done is Taylor expansion for $e^z$ $$e^z = \sum_{k=0}^\infty \frac{z^k}{k!}$$ And $\frac{1}{z}$ substitution $$e^\frac{1}{z} = \sum_{k=0}^\infty \frac{1}{k!} \frac{1}{z^k}$$ But this substitution leads to expansion about $z=\infty$
How can I get expansion about $z=0$?

Answer 1

But you did get the Laurent series about $0$:$$1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}\cdots$$