Proving that : $ \frac{W(x)}{xe^x}=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!}T(n)x^n $ How to prove that: 
$$ \frac{W(x)}{xe^x}=\sum_{n=0}^{\infty} \frac{(-1)^n}{n!}T(n)x^n $$
where 
$T(n)$
counts the number of forests of rooted labeled trees using labels in a subset of 
$\{1,\ldots,n\}$ and $W(x)$ is the Lambert $W$ function?
Also, as shown in this video at 6:54, does $T (n)=n^{n-2} $ ?
 A: This is not an answer.
This is a problem of Taylor series composition using
$$W(x)=\sum_{n=1}^\infty (-1)^n\,\frac{n^{(n-1)}}{n!}x^n$$ (have a look here)
$$e^{-x}=\sum_{n=0}^\infty (-1)^n\,\frac{x^n}{n!}$$ So, computing the first terms, we have, as a series,
$$\frac{W(x)}{xe^x}=1-2 x+3 x^2-\frac{29 x^3}{6}+\frac{53 x^4}{6}-\frac{2117 x^5}{120}+\frac{2683
   x^6}{72}-\frac{82403 x^7}{1008}+O\left(x^8\right)$$ making the $T_n$ to be the sequence
$$\{1,2,6,29,212,2117,26830,412015\}$$ which is sequence $A088957$ in $OEIS$.
In a comment, Alex Chin wrote that these coefficients are "the number of rooted subtrees (for a fixed root) in the complete graph on $n$ vertices" (please : do not ask me what this means !).
A: Start   by  making   some   definitions.    The  combinatorial   class
$\mathcal{T}$ of labelled trees has the specification
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} = 
\mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
which gives the functional equation
$$T(z) = z \exp T(z).$$
We   have   by   Cayley   that
$$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}.$$
We seek to prove that
$$\bbox[5px,border:2px solid #00A000]{
W(z) = z \exp(z) \sum_{n\ge 0} (-1)^{n} Q_n 
\frac{z^n}{n!}}$$
where $Q_n$ is the number of rooted  subtrees for a fixed root in the
complete graph $K_{n+1}$  on $n+1$ vertices. These have  from $k=0$ to
$k=n$ nodes not  including the fixed root and represent  a forest that
is   attached    to   the   root.    (Specification    as   in   OEIS
A088957,  example given  by  A. Chin.)   The
combinatorial class $\mathcal{F}$ of forests is given by
$$\mathcal{F} = \textsc{SET}(\mathcal{T})$$
and hence has EGF
$$F(z) = \exp T(z) = \frac{1}{z} T(z).$$
We also have by construction that
$$Q_n = \sum_{k=0}^{n} {n\choose k}
k! [z^k] F(z).$$
Here we chose  the labels that go into the  forest and substitute them
into the  forest respecting the  ordering of  the nodes in  the source
forest. It follows by convolution of  EGFs that (recall that $n! [z^n]
\exp(z) = 1$)
$$Q(z) = \sum_{n\ge 0} Q_n \frac{z^n}{n!}
= \exp(z) \frac{1}{z} T(z).$$
Now by definition the principal branch of the Lambert W function
is 
$$W(z) = \sum_{n\ge 1} (-1)^{n-1} n^{n-1} \frac{z^n}{n!}$$
and hence we obtain
$$\bbox[5px,border:2px solid #00A000]{
\sum_{n\ge 0} Q_n 
\frac{z^n}{n!}  = \frac{1}{z} T(z) \exp(z)
= - \frac{1}{z} W(-z) \exp(z).}$$
The claim now follows by replacing $z$ by $-z$ to get
$$\sum_{n\ge 0} (-1)^n Q_n 
\frac{z^n}{n!}  = \frac{1}{z} W(z) \exp(-z).$$
