Note: throughout this question, I'll be using the following notation convention:
$$f^{(n)}(x)=\frac{d^nf}{dx^n}(x)$$
I was browsing through Wikipedia and even MSE's related questions searching for a proof for the Lagrange Inversion Theorem. I'm letting the function $g$ be $f$'s inverse, and I'm letting $f$ be analytic at $a$ while also maintaining the property that $f'(a) \neq 0$.
$g$ is also assumed to be analytic, therefore suggesting that
$$g(f(w))=\displaystyle\sum_{n=0}^{\infty}g^{(n)}(f(a))\frac{(f(w)-f(a))^n}{n!}=w.$$
Expanding past the zeroth term, we get
$$g(f(w))=a+\displaystyle\sum_{n=1}^{\infty}g^{(n)}(f(a))\frac{(f(w)-f(a))^n}{n!}.$$
Now, letting $f(w)=z$, we have
$$g(z)=a+\displaystyle\sum_{n=1}^{\infty}g^{(n)}(f(a))\frac{(z-f(a))^n}{n!}.$$
Wikipedia states that
$$g(z)=a+\displaystyle\sum_{n=1}^{\infty}\lim_{w \to a}\Bigg[\frac{d^{n-1}}{dw^{n-1}}\Bigg(\frac{w-a}{f(w)-f(a)}\Bigg)^n\Bigg]\frac{(z-f(a))^n}{n!},$$
which suggests the following:
$$g^{(n)}(f(a))=\lim_{w \to a}\Bigg[\frac{d^{n-1}}{dw^{n-1}}\Bigg(\frac{w-a}{f(w)-f(a)}\Bigg)^n\Bigg].$$
It looks like a residue of some weird function at $a$, but the nature of the function itself and the reason why it shows up from a derivative rather than an integral are really stumping me. I've read a few pdf's related to it, but one gave an answer that was too far from the equation itself for me to draw a connection, and the other was in French (my high school French classes have failed me, needless to say).
This is primarily just me wanting to find an inverse polylogarithm, a question that has already been asked here on MSE with no avail, but also for me to understand this theorem more precisely.
If someone could please point out where this residue-ish thing comes from and why it looks the way it does (i.e., no factorial or $2\pi i$ term), I'd be greatly appreciative. If instead you were able to give me a derivation or a proof of the polynomial approximation given by the commenter Simon here, that would be great too. A few more things that may or may not be useful specifically for the polylogarithms are the following equations:
$$z _{s+1}F_s(1,\dots,1;2,\dots,2;z)=\operatorname{Li}_s(z)$$
$$(z-1) _2F_1(1,1;2;1-z)=\log(z)$$
$$\operatorname{Li}_1(z)=-\log(1-z)$$
$$\operatorname{Li}_0(z)=\frac{z}{1-z}$$
$$\operatorname{Li}_1^{-1}(z)=1-e^{-z}$$
$$\operatorname{Li}_0^{-1}(z)=\frac{z}{1+z}$$
were $_pF_q(a_1,\dots,a_p;b_1,\dots,b_q;z)$ denotes the generalized hypergeometric function and $\log$ denotes the natural logarithm.