How to derive the Lambert W function series expansion? How do you use the Lagrange inversion theorem to derive the Taylor Series expansion of W(x)? How else can you derive a series expansion?
There is  a variant of  Lagrange inversion which  I like to  call poor
man's Lagrange  inversion which consists  in using the  Cauchy Residue
Theorem.

In the present case we are looking for the inverse to $ze^z$ so that
$$W(x) e^{W(x)} = x.$$
From the Cauchy Residue Theorem we get
$$[x^n] W(x)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} W(z) \; dz.$$
Now put $W(z) = w$  so that $w e^w = z$ and $e^w (1+w)  \; dw = dz$ to
obtain (we use the branch that  takes $w=0$ to $z=0$, there is another
that takes $w=-\infty$ to $z=0$)
$$\frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{1}{w^{n+1}} e^{-w(n+1)} e^w (w+w^2) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma} 
\frac{1}{w^{n+1}} e^{-wn} (w+w^2) \; dw.$$
Extracting coefficients now yields
$$\frac{(-1)^{n-1} n^{n-1}}{(n-1)!}
+ \frac{(-1)^{n-2} n^{n-2}}{(n-2)!}
\\ = \frac{(-1)^{n-1}}{n!} n^{n-1}
(n - (n-1)) = \frac{(-1)^{n-1}}{n!} n^{n-1}.$$
We conclude that $$W(x) = \sum_{n\ge 1}
\frac{(-1)^{n-1}}{n!} n^{n-1} x^n.$$
Here is an example I at MSE and another example II at MSE of Lagrange inversion.
Addendum, Sep. 17 2020. For a more compact version use Lagrange
Inversion as presented in Analytic  Combinatorics by Flajolet and
Sedgewick. We first choose the branch  that has $W(0) = 0.$ We then have
$W'(x) e^{W(x)} + W(x) e^{W(x)}  W'(x) = 1$ so that $W'(1) = 1$. This
means we may write for $n\ge 1$ by  the Cauchy Coefficient Formula
$$[x^n] W(x) = \frac{1}{n} [x^{n-1}] W(x) =
\frac{1}{n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^n} W'(z) \; dz.$$
Putting $W(z) = w$ we also have $W(z) = z + \cdots$ so that the image  of
the circle in $z$ is a contour in $w$ that makes one turn by the  origin
and may in  turn be deformed to another small circle. We then have
$$[x^n] W(x) =
\frac{1}{n} \frac{1}{2\pi i}
\int_{|w|=\gamma} \frac{1}{w^n} \exp(-nw) \; dw.$$
This is by inspection
$$\frac{1}{n} [w^{n-1}] \exp(-nw) =
\frac{1}{n} \frac{(-n)^{n-1}}{(n-1)!} =
\frac{(-1)^{n-1}}{n!} n^{n-1}$$
as before.
It can be easily derived by applying the Lagrange inversion formula to $ze^z$ as it satisfies all criterias of the Lagrange inversion theorem, i.e. $f(0)=0$ and $f'(0)$ is non zero.