the inverse of $f(x)=x-x^p$ 
Provided that $f(x)=x-x^p$, prove that $$f^{-1}(x)=\sum_{k\ge1}{pk\choose k}\frac{x^{1+(p-1)k}}{1+(p-1)k}.$$

I have gotten so far as using the Lagrange Inversion Theorem to show that
$$f^{-1}(x)=\sum_{k\ge1}g_k\frac{x^k}{k!}$$
where 
$$g_k=\lim_{w\to0}\left[\left(\frac{d}{dw}\right)^{k-1}(1-w^{p-1})^{-k}\right],$$
but I have no idea how to compute this limit, or where to go from there. Could I have some help? Thanks :)
 A: As written, this works only for integer $p>1$. You can use (binomial expansion)
$$(1-w^{p-1})^{-k}=\sum_{n\geqslant 0}\binom{n+k-1}{n}w^{n(p-1)},$$
see that $g_k\neq 0$ only when $k-1=n(p-1)$ for an integer $\color{red}{n\geqslant 0}$, and get
$$g_{n(p-1)+1}=\binom{np}{n}\big(n(p-1)\big)!$$
as expected. But there is a simplification that also handles non-integer $p>1$. The solution $w(z)$ of $w-w^p=z$ we look for is of the form $w(z)=zy(z)$, where $y-z^{p-1}y^p=1$, so that $y$, as a function of $x=z^{p-1}$, satisfies $$x=\frac{y-1}{y^p},$$
and we apply the theorem to this equation (at $y=1$). This gives $y=1+\sum\limits_{k\geqslant 1}g_k x^k$, where
$$g_k=\frac{1}{k!}\lim_{y\to 1}\left(\frac{d}{dy}\right)^{k-1}\color{blue}{y^{pk}}=\frac{1}{k}\binom{pk}{k-1}=\frac{1}{(p-1)k+1}\binom{pk}{k}.$$
A: Suppose we have
$$z = q(z) - q(z)^p$$
with $p\ge 2$ an integer and we seek
$$q(z) = \sum_{n\ge 0} Q_n z^n.$$
Start with some basic observations namely that
$$[z^0] (q(z)-q(z)^p) = [z^0] z = 0 = Q_0 - Q_0^p.$$
We will choose the branch that has $Q_0 = 0.$
Furthermore we have
$$[z^1] (q(z)-q(z)^p) = [z^1] z = 1 =
[z^1] (Q_1 z + \cdots - Q_1^p z^p - \cdots) = Q_1$$
and hence $Q_1 = 1.$ Using the Cauchy Coefficient Formula
we write
$$n Q_n = [z^{n-1}] q'(z) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} q'(z) \; dz.$$
We put $q(z) = w$ to that $q'(z) \; dz = dw.$ With the chosen branch
we obtain
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{(w-w^p)^n} \; dw
= \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^n} \frac{1}{(1-w^{p-1})^n} \; dw.$$
This yields with the factor in front
$$\frac{1}{n} [w^{n-1}] \frac{1}{(1-w^{p-1})^n}$$
so that we must have $n=(p-1)k+1$ where $k\ge 0.$ We find
$$\frac{1}{(p-1)k+1}
[w^{(p-1)k}] \frac{1}{(1-w^{p-1})^{(p-1)k+1}}
\\= \frac{1}{(p-1)k+1}
[w^{k}] \frac{1}{(1-w)^{(p-1)k+1}}
= \frac{1}{(p-1)k+1} {k+(p-1)k\choose k}.$$
This finally yields
$$\bbox[5px,border:2px solid #00A000]{
q(z) = \sum_{k\ge 0} \frac{z^{(p-1)k+1}}{(p-1)k+1}
{pk\choose k}.}$$
