How one deals with series like this one: $zb + \frac{z^2}{2!}b^2(1+\frac{c}{b}) + \frac{z^3}{3!}b^3(1+\frac{2c}{b})(1+\frac{c}{b}) + \frac{z^4}{4!}b^4(1+\frac{3c}{b})(1+\frac{2c}{b})(1+\frac{c}{b})+...$?
The coefficients' complexity increases but in a highly "predictable" way.
In terms of a hypergeometric function, your series is$$\sum_{z\ge1}\left(\frac{(cz)^k}{k!}\frac1b\prod_{j=0}^{k-1}\left(\frac{b}{c}+j\right)\right)=\frac{{}_1F_0(\frac{b}{c};\,cz)-1}{b}.$$