Let $f\colon G\rightarrow\mathbb{C}$ be a complex valued function given by $f(z)=\exp(\frac{z}{1-z})$.
Prove that the Taylor series' coefficients of $f$ at $0$ are $$a_0=1 \qquad \qquad a_n=\sum_{s=1}^n \frac{1}{s!} \binom{n-1}{s-1}$$
Thoughts: My idea was to do induction in n. I found $a_0=1$ pretty easy as $\exp(0)=1$. It's the induction step that is causing me some trouble and I need some help with! Also is there an easier way than induction to prove this? Thank you!
Notice that \begin{align*} \exp\Big(\frac z{z-1}\Big) &= \sum_{k=0}^\infty \frac1{k!}\Big(\frac{z}{1-z}\Big)^k\\[3pt] &= \sum_{k=0}^\infty \frac{z^k}{k!}(1-z)^{-k}\\[3pt] &= 1+ \sum_{k=1}^\infty \frac{1}{k!}\Big(\sum_{n=0}^\infty\frac{(k+n-1)!}{(k-1)!\,n!}\, z^{n+k}\Big)\\[3pt] &= 1+\sum_{k=1}^\infty \frac{1}{k!}\Big(\sum_{n=k}^\infty\frac{(n-1)!}{(k-1)!\,(n-k)!}\, z^n\Big) \quad[n\leftarrow n-k]\\[3pt] &= 1+\sum_{k=1}^\infty \frac{1}{k!}\Big(\sum_{n=k}^\infty\binom{n-1}{k-1} z^n\Big)\\[3pt] &=1+\sum_{k=1}^\infty \sum_{n=k}^\infty\frac{1}{k!}\binom{n-1}{k-1} z^n\\[3pt] &=1+\sum_{n=1}^\infty \Big(\sum_{s=1}^n\frac{1}{s!}\binom{n-1}{s-1}\Big) z^n, \end{align*} where the third step applies the binomial theorem to $(1-z)^{-k}$, and the last step follows by swapping the order of summation.