How to prove that, if $$ w=ze^{aw^m} \quad (m\in\mathbb{N}), $$ then $$ w=\sum_{n=0}^\infty\frac{a^n(nm+1)^{n-1}} {n!}z^{nm+1} \; ? $$
I am aware of the Bürmann-Lagrange inversion formula, that if $z=we^{-aw^m},$ then $$ w=\sum_{n=0}^\infty\frac{z^{n+1}}{(n+1)!} \lim_{z\to0}\frac{d^n e^{a(n+1)z^m}}{dz^n}, $$ But I have difficulty proving that $$ \lim_{z\to0}\frac{d^n e^{a(n+1)z^m}}{dz^n}= \begin{cases} a^k(km+1)^nkm\cdots(n+1),& \text{if } \; n=km+1,\\ 0, & \text{otherwise}. \end{cases} $$ There should probably be a better way to establish the expansion.
The method for Lagrange inversion by the Cauchy Coefficient Formula applies here, as documented and proved on page $732$, section $A.6$ of Analytic Combinatorics by Flajolet and Sedgewick. The reader is invited to consult the text for additional information. Introducing
$$Q(z) = \sum_{n\ge 1} Q_n z^n$$
with
$$Q(z) = z \exp(a Q(z)^m)$$
we follow the given reference in writing
$$n Q_n = [z^{n-1}] Q'(z) = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^n} Q'(z) \; dz.$$
Now with $z = Q(z) \exp(-a Q(z)^m)$ this is
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{Q(z)^n} \exp(a n Q(z)^m) Q'(z) \; dz.$$
Put $Q(z) = w$ to get
$$\frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^n} \exp(a n w^m) \; dw = [w^{n-1}] \exp(a n w^m).$$
Here we must have $n=pm+1$ with $p\ge 0$ and we obtain
$$[w^{pm}] \exp(a (pm+1) w^m) = [w^p] \exp(a (pm+1) w) = \frac{a^p (pm+1)^p}{p!}.$$
We thus have
$$Q_n = \frac{1}{pm+1} \frac{a^p (pm+1)^p}{p!} z^{pm+1}$$
or
$$\bbox[5px,border:2px solid #00A000]{ Q(z) = \sum_{p\ge 0} \frac{a^p (pm+1)^{p-1}}{p!} z^{pm+1}.}$$