Disclaimer: This is a homework problem for my graduate Applied Complex Variables course. The book being used is "Applied and Computational Complex Analysis: Volume 1" by Peter Henrici.
Problem:
Let $a,b,c$ be complex numbers, $a\neq b,c\neq 0$, and let
\begin{align*}
w=f(z):=\frac{1}{c}(e^{az}-e^{bz}).
\end{align*}
Show that the inverse function near $w=0$ is represented by
\begin{align*}
z=f^{[-1]}(w)=\frac{1}{d}\sum_{n=1}^\infty \frac{(-1)^{n-1}c^n}{n!}\left(\frac{nb}{d}+1\right)_{n-1}w^n,
\end{align*}
where $d:=a-b$. Derive as special cases the representations for the inverses of the functions
\begin{align*}
w=\sin z,\quad w=e^z-1,\quad w=e^{\alpha z}\sin \beta z,
\end{align*}
and, as the limiting case $a=b+c$, $c\to 0$, $w=ze^{bz}$. [To compute residues use formal integration by parts; see Problem $1.8.6$.]
Attempted Solution: Since $e^z$ is analytic at zero it follows that $\frac{1}{c}(e^{az}-e^{bz})$ is analytic and has a power series representation near $z=0$. Representing $f(z)$ by its power series it follows: $$w=f(z)=\frac{1}{c}(e^{az}-e^{bz})=\frac{1}{c}\sum_{n=1}^\infty\frac{a^n-b^n}{n!}z^n=F.$$ Applying Lagrange Inversion to $H:=[F]_1^\frac{1}{n}Z(B_\frac{1}{n}\circ Q)$ where $$Q:=\sum_{n=1}^\infty\frac{a^{n+1}-b^{n+1}}{(n+1)!(a-b)}z^n$$ we have the following: $$z=H^{[-1]}(w)=\sum_{m=1}^\infty\frac{C^m}{m(a-b)^m}\text{res}[Z^{-m}(B_{-m}\circ Q)]z^m.$$ The residue is then the $(m-1)$st coefficient of $B_{-m}\circ Q$ which can be calculated by J.C.P. Miller which gives $$[B_{-m}\circ Q]_0=1, \quad [B_{-m}\circ Q]_{m-1}=\frac{1}{m-1}\sum_{n=1}^{m-1}[(-m+1)k-(m-1)]\frac{a^{k+1}-b^{k+1}}{(k+1)!(a-b)}[B_{-m}\circ Q]_{m-1-k}.$$ Some simplification can be done to the second expression to get $$[B_{-m}\circ Q]_{m-1}=-\sum_{n=1}^{m-1}\frac{a^{k+1}-b^{k+1}}{k!(a-b)}[B_{-m}\circ Q]_{m-1-k}.$$ However, I am uncertain as to where to go from here. The question suggests using the identity $$\text{res}(M'N)=-\text{res}(N'M),$$ where $M$ and $N$ are formal Laurent series, and formal integration by parts to calculate the residue. It is possible to try and find $F^{[-1]}$ directly, though it would be a far more difficult calculation. Thank you in advance for any insight you may be willing to provide.
Update: I have reduced the problem to showing the following: $$\text{res}[d^{-n}Z^{-n}(B_{-n}\circ Q)]=\binom{\frac{nb}{d}+1}{n-1}$$ This was done by analyzing the desired result and form of $z$.