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?
[Math] How to derive the Lambert W function series expansion
lambert-w
Related Solutions
Use the MATHEMATICA commands
f = -E/(2 z) ProductLog[-2 z/E^2] (Log [-E/(2 z) ProductLog[-2 z/E^2]] - 1)
Series[f, {z, 1, 10}]//N
and you will obtain
$$ f(z)=-0.880194-0.22445 (z-1.)-0.153651 (z-1.)^2-0.164248 (z-1.)^3-0.224172 (z-1.)^4-0.355448 (z-1.)^5-0.622729 (z-1.)^6-1.17077 (z-1.)^7-2.31929 (z-1.)^8-4.78304 (z-1.)^9-10.1829 (z-1.)^{10}+O\left((z-1.)^{11}\right) $$
or if you prefer the lenghty form (only four therms)
$$ f(z) = \frac{1}{2} e W\left(-\frac{2}{e^2}\right) \left(W\left(-\frac{2}{e^2}\right)+2\right)-\frac{1}{2} \left(e W\left(-\frac{2}{e^2}\right)^2\right) (z-1)+\frac{e W\left(-\frac{2}{e^2}\right)^3 (z-1)^2}{2 W\left(-\frac{2}{e^2}\right)+2}-\frac{\left(e W\left(-\frac{2}{e^2}\right)^4 \left(3 W\left(-\frac{2}{e^2}\right)+4\right)\right) (z-1)^3}{6 \left(W\left(-\frac{2}{e^2}\right)+1\right)^3}+\frac{e W\left(-\frac{2}{e^2}\right)^5 \left(2 W\left(-\frac{2}{e^2}\right) \left(6 W\left(-\frac{2}{e^2}\right)+17\right)+25\right) (z-1)^4}{24 \left(W\left(-\frac{2}{e^2}\right)+1\right)^5}+O\left((z-1)^5\right) $$
etc.
You can also use the fact
$$ f(z) = y(z)(\ln y(z) -1) $$
with
$$ y(z) = -\frac{e W\left(-\frac{2 z}{e^2}\right)}{2 z} $$
and
$$ y_0 = y(1) = -\frac{1}{2} e W\left(-\frac{2}{e^2}\right) $$
and then find the expansion for $y$. This can be done with MATHEMATICA or with bare hand.
This answer does not fully address the question of how to compute an explicit formula from the recursion relation (though it indicates a reference which contains a starting point), and so is not a proper answer, but it is too long for a comment, and I expect it might still be useful.
It follows from a comment by Vladimir Kruchinin in OEIS A042977, "Triangle $T(n,k)$ read by rows: coefficients of a polynomial sequence occurring when calculating the $n$th derivative of Lambert function $W$" that $$\color{#df0000}{\boxed{P_n(w) = \sum_{m = 0}^n \left[ \sum_{j=0}^m {2 n + 1 \choose m - j} \sum_{k=0}^j (-1)^k \frac{(n+k+1)^{n+j}}{(j-k)! k!} \right] w^m}} .$$ See under the heading $\texttt{FORMULA}$, and NB that the expression in this answer differs from the one in OEIS entry by a factor of $(-1)^n$ to account for the formulation in the question. This expression can be deduced from Example 4.3 in Kruchinin's preprint "Derivation of Bell Polynomials of the Second Kind".
In particular, to compute the constant term of $P_n(w)$, we can evaluate the inner sum for $j = 0$. Both summations have only a single term, and it simplifies to the conjectured formula, $$\color{#df0000}{\boxed{P_n(0) = (n + 1)^n}} .$$
Best Answer
While the common way to derive it is by using the Lagrange Inverse Theorem, there technically isn't anything stopping us from making a Taylor Series for it as you would with any other function. As always, we're going to need a list of derivatives. The first one can be found pretty easily via implicit differentiation as follows:
$$y = W(x)$$ $$ye^y = x$$ $$\frac{d}{dx}\left(ye^y=x\right)$$ $$\left(y+1\right)e^y\cdot\frac{dy}{dx}=1$$ $$\therefore \frac{d}{dx}W\left(x\right)=\frac{e^{-W\left(x\right)}}{\left(W\left(x\right)+1\right)}$$ $$\frac{d}{dx}W\left(x\right)=\frac{W\left(x\right)}{x\left(W\left(x\right)+1\right)}$$
Also, to anyone unfamiliar with the last step, it is one of the main identities of the Lambert W function.
Now that we have the first derivative, we can simply differentiate as many times as we want to get all subsequent derivatives. The important thing to note, however, is that all subsequent derivatives will only require the W function to be evaluated at x. Therefore, if we know the value of W(x) we can theoretically calculate the value of any nth derivative of W(x) at that x. We can use this to our advantage by considering a value of W which is easy to calculate, such as W(e). This can be calculated as follows:
$$y=W\left(e\right)$$ $$ye^y=e=1e^1$$ $$\therefore y=1 \Rightarrow W(e)=1 $$
We now have everything we need to calculate a Taylor Series centered at x=e as usual.
$$\sum_{n=0}^{\infty}\frac{W^{\left(n\right)}\left(e\right)}{n!}\left(x-e\right)^n$$
$$=1+\frac{1\left(x-e\right)}{2\cdot e\cdot1!}-\frac{3\left(x-e\right)^2}{2^3e^2\cdot2!}+\frac{19\left(x-e\right)^3}{2^5e^3\cdot3!}-\frac{185\left(x-e\right)^4}{2^7e^4\cdot4!}+\frac{2437\left(x-e\right)^5}{2^9e^5\cdot5!}...$$
I agree that this isn't as useful as the Lagrange Inverse method (mostly due to the lack of an explicit definition for the coefficients), however I felt it was worth noting as the question did ask if there were any other methods. And, as shown by the plot of the 5th degree polynomial, this does indeed work.
5th Degree Polynomial Approximation of W(x)