Lagrange inversion theorem applied on $f(x)=x-x^p$

Question

I was looking at the Wikipedia page of the Lagrange inversion theorem and stumbled upon the example. It says that to find the root of the polynomial $x^p-x+z=0$ we can find the inverse of the function $f(x)=x-x^p$ using the theorem. So we have $f(x)=z$, therefore the inverse is $x=g(z)$ where $$ g(z)=x_0+\sum_{n=1}^\infty \frac{g_n}{n!}(z-f(x_0))^n \qquad g_n=\lim_{w\to x_0}\frac{\partial^{n-1}}{\partial w^{n-1}}\left[\left(\frac{w-x_0}{f(w)-f(x_0)}\right)^n\right] $$ The result should be, according to Wikipedia: $$ g(z)=\sum_{n=1}^\infty\binom{pk}{k}\frac{z^{(p-1)k+1}}{(p-1)k+1} $$ So, I decided to try it myself. First I put $x_0=0$, because $f'(0)=1$, and calculated $g_n$ $$ g_n=\lim_{w\to0}\frac{\partial^{n-1}}{\partial w^{n-1}}\left[(1-w^{p-1})^{-n}\right] $$ Using the negative binomial theorem we obtain $$ g_n=\lim_{w\to0}\frac{\partial^{n-1}}{\partial w^{n-1}}\left[\sum_{k=0}^\infty(-1)^k\binom{n+k-1}{k}w^{-(n+k)(p-1)}\right]= \qquad |w|<1\\ =\lim_{w\to0}\sum_{k=0}^\infty(-1)^k\binom{n+k-1}{k}(-(n+k)(p-1))...(-(n+k)(p-1)-n-2)w^{-(n+k)(p-1)-n-1} $$ Now, this implies that $g_n=0$ $\forall n\geq2$, which isn't right. What did I do wrong? Thanks.