$W_{-1} (x)$ series expansion? By the Lagrange Inversion Theorem, one can derive the series expansion for the principal branch of $W_0(x)$:
$$W_0(x)= \sum_{n=1}^{\infty} \frac{(-n)^{n-1}x^n}{n!}, \, |x| \leq \frac1e$$
For $x \in \mathbb{R}$.
Is there also a series expansion for $W_{-1}(x)$? If so, how can it be derived?
 A: Here are $2$ series expansions:
$$\text W_{-1}(z)=\ln(z)-2i\pi-\ln(\ln(z)-2i\pi)-\sum_{n=1}^\infty\sum_{m=1}^n\frac{\text S_n^{(m-n+1)}\ln^m(\ln(z)-2i\pi)}{(2\pi i-\ln(z))^nm!} $$
from:
The Generalized Lambert function from Wolfram Functions and the Stirling numbers of the first kind
and
$$-Q^{-1}(2,-ez)-1\mathop=^{-\frac1e\le z<0}\text W_{-1}(z)=…-\frac{43}{135}(ex+1)^2-\frac{11}{72}(2ex+2)^\frac32-\frac23(ex+1)-\sqrt{2ex+2}-1 $$

From the series expansion of Inverse Gamma Regularized. Please correct me and give me feedback!
Is there any pattern for the bottom series?
A: Let $y=xe^x$; so $x=W_{-1}(y)$, where $y\to 0^-$ as $x\to-\infty$.  Note
that $y=0$ is a logarithmic branch point for $W_{-1}(y)$.  The series for $x$ in terms of $y$ presumably begins like this:
$$
x = \log(-y) - \log(-\log(-y)) + \dots\qquad\text{as } y \to -\infty
$$

Derivation ... I prefer asymptotics for $X \to +\infty$, so write
$X=-x$ and $Y = -1/y$.  So $X \to +\infty$ and $Y\to + \infty$, with
$$
e^X = XY,
$$
$$
X = \log X + \log Y
\tag0$$
Now $\log X= o(X)$ so from $(0)$ we get
$$
X = \log Y + o(\log Y)
\tag1$$
That is the first approximation.
Next, compute
$$
\log X = \log(\log Y + \log X)
= \log\left((\log Y)\left(1+\frac{\log X}{\log Y}\right)\right)
= \log\log Y + \log\left(1+\frac{\log X}{\log Y}\right)
$$
Thus
$$
X = \log Y + \log\log Y + \log\left(1+\frac{\log X}{\log Y}\right)
$$
Here, $\log\log Y \to +\infty$, but $\log X/\log Y \to 0$ so
$\log\left(1+\frac{\log X}{\log Y}\right) \to 0$, so
$$
X = \log Y + \log\log Y + o(1)
\tag2$$
That is the second approximation.
We could continue this to get as many terms as we want.
The next term is not $\log\log\log Y$.  I think the next one is
$$
X = \log Y + \log\log Y + \frac{\log\log Y}{\log Y} + \dots
\tag3$$
Substitute in $(2)$ for $x, y$ to get
$$
W_{-1}(y) = x = \log(-y) - \log(-\log(-y)) + o(1)  \qquad\text{as } y \to -\infty
$$
