Bürmann-Lagrange formula for a function 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.
 A: 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
$$\bbox[5px,border:2px solid #00A000]{
Q(z) = \sum_{p\ge 0}
\frac{a^p (pm+1)^{p-1}}{p!} z^{pm+1}.}$$
