For analytic $f$, how can I represent the expression $f(z)\cdot\exp\left({s\,\log(z)}\right)$, i.e. $f(z)\cdot z^s$ in the form
$$\sum_{n}^\infty\left(\sum_{k}^\infty a_k s^k\right)z^n,$$
at least as a formal power series, where the index runs over the number needed?
I got to $$f(z)\cdot\exp\left({s\,\mathrm{log}(z)}\right)$$ $$=\left(\sum_{m}^\infty\frac{1}{m!}f^{(m)}(0)\,z^m\right) \left(\sum_{j=0}^\infty \frac{1}{j!} \left(s\,\log{(z)}\right)^j\right)$$ $$=\sum_{m}^\infty\sum_{j=0}^\infty \frac{1}{m!\,j!}f^{(m)}(0)\,s^j\cdot z^m\, \log{(z)}^j,$$
and I know
$$\log(z)=\sum_{l=1}^\infty (-1)^{l+1}\frac{1}{l}(z-1)^{l}.$$
I'm motivated by wanting to understand the Mellin transform (and the Laplace transform, for that matter) and here I approach this by looking at what it does to series components of a function
$$f(z)=\sum c_n z^n\mapsto \mathcal M(f)=\sum c_n^\mathcal{M} z^n$$
