We start with the combinatorial class
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\mathcal{T} =
\mathcal{Z} \times \mathcal{Y} +
\mathcal{Z}\times \textsc{SET}_{\ge 1}(\mathcal{T}).$$
This translates into the functional equation
$$T(z) = zy + z\times(\exp T(z) - 1) = z\times(-1+y+\exp T(z))$$
or
$$z = \frac{T(z)}{-1+y+\exp T(z)}.$$
Writing
$$T(z) = \sum_{n\ge 1} T_n(y) \frac{z^n}{n!}$$
we are interested in extracting coefficients as per the residue
operator
$$\frac{1}{(n-1)!} T_n(y) = [z^{n-1}] T'(z) =
\; \underset{z}{\mathrm{res}} \; \frac{1}{z^{n}} T'(z).$$
Now we put $w=T(z)$ so that $dw = T'(z) \; dz$ and use the functional
equation to obtain
$$\; \underset{w}{\mathrm{res}} \;\frac{(-1+y+\exp w)^n}{w^{n}}$$
Extracting the coefficient on $[y^k]$ we find
$$[y^k] \; \underset{w}{\mathrm{res}}
\;\frac{(-1+y+\exp w)^n}{w^{n}}
= {n\choose k} \; \underset{w}{\mathrm{res}} \;
\frac{(\exp(w)-1)^{n-k}}{w^{n}}.$$
This yields
$$[y^k] T_n(y) = (n-1)! {n\choose k}
[w^{n-1}] (\exp(w)-1)^{n-k}
\\ = \frac{n!}{k!}
(n-1)! [w^{n-1}] \frac{(\exp(w)-1)^{n-k}}{(n-k)!}.$$
We recognize the EGF of the Stirling numbers of the second kind,
getting
$$\bbox[5px,border:2px solid #00A000]{
\frac{n!}{k!} {n-1\brace n-k}.}$$
This is the residue operator from Egorychev's Combinatorial sums.
As a sanity check we find for the number of all rooted trees
$$\sum_{k=1}^n
(n-1)! {n\choose k}
[w^{n-1}] (\exp(w)-1)^{n-k}
\\ = (n-1)! [w^{n-1}]
\sum_{k=1}^n {n\choose k} (\exp(w)-1)^{n-k}.$$
Now for $k=0$ we get
$$(n-1)! [w^{n-1}] (\exp(w)-1)^n = 0$$
because $\exp(w)-1 = w + \cdots.$ Hence we may continue with
$$(n-1)! [w^{n-1}]
\sum_{k=0}^n {n\choose k} (\exp(w)-1)^{n-k}
= (n-1)! [w^{n-1}] \exp(nw) \\ = (n-1)! \frac{n^{n-1}}{(n-1)!}
= n^{n-1}$$
and the check goes through (Cayley's formula).
