Expand $e^{\frac{1}{z-1}}$ in a Laurent series for $|z|>1$ We have :$$e^{\frac{1}{z-1}}=e^{\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}}$$
So $$e^{\frac{1}{z-1}}= 1 +(1+z+z^2+...)+\frac{1}{2}(1+z+z^2+...)^2 +...$$
The result is $$e^{\frac{1}{z-1}}=\sum_{n=0}^{\infty}\frac{a_n}{z^n}$$ Where $$a_n = \sum_{k=1}^{\infty}\binom{n-1}{k-1} \frac{1}{k!}$$
Edit : $$a_n = \sum_{k=1}^{n}\binom{n-1}{k-1} \frac{1}{k!}$$ Is there a trick to find the result without a lot of calculation ?
Thank you ...