Getting the proof of the generating function formula for Stirling numbers by 'staring' at the expression using combinatorial classes. In his lectures on enumerative combinatorics, Prof. Federico Ardila gives a homework. For a fixed natural $k$, he asks us to prove the identity: 
$$\sum_{n \geq 0} S(n,k) z^n = \left(\frac{z}{1-z}\right)\left(\frac{z}{1-2z}\right)...\left(\frac{z}{1-kz}\right).$$
Here, $S(n,k)$ is the number of ways of partitioning a $n$-element set into $k$ sets.
I know the proof of the above identity using recursions. I also know combinatorial proofs of related identities. But I want to know the proof using combinatorial classes.
It looks like the combinatorial class of $k$-partitions is the product of the classes Seq({$0,1,2,...,l$}) from $l=0$ to $l=k-1$. But I am unable to find an such an interpretation. 
How do I solve this problem? 
Thanks.
 A: The difficulty here  is that you have an OGF  where most commonly EGFs
are  used. Nonetheless a  proof using  unlabeled species  and advanced
Polya Enumeration can be given.

Some time  ago I posted  an intriguing result  proving the EGF  of the
Stirling numbers ${n\brace k}$ with  $k$ fixed which seems not to have
found an audience. This is the
MSE link. 
  We  can  use  the  material from  this  link  without  explicitly
recapitulating everything  as it is  sound. The following  formula was
proved there by methods of species theory (unlabeled):
$${n\brace k} = [w^k] e^{-w}
\left.\left(v\frac{d}{dv}\right)^n e^{vw} \right|_{v=1}.$$
Introduce the generating function
$$P(z) = \sum_{n\ge 0} {n\brace k} z^n
= [w^k] e^{-w}
\left.\sum_{n\ge 0} z^n
\left(v\frac{d}{dv}\right)^n e^{vw} \right|_{v=1}.$$
Now the operator represented by the sum turns $v^q w^m$ into
$$\left(1 + (zq)
+ (zq)^2
+ (zq)^3
+ \cdots\right) \times v^q \times w^m
= \frac{1}{1-qz} \times v^q \times w^m.$$
Therefore $$e^{vw} = \sum_{q\ge 0} \frac{v^q w^q}{q!}$$ is transformed
into $$\sum_{q\ge 0} \frac{1}{1-qz} \frac{v^q w^q}{q!}.$$
This yields for $P(z)$ that
$$P(z) = [w^k] e^{-w}
\left. \sum_{q\ge 0} \frac{1}{1-qz} \frac{v^q w^q}{q!} \right|_{v=1}
= [w^k] e^{-w}
\sum_{q\ge 0} \frac{1}{1-qz} \frac{w^q}{q!}.$$
Actually doing the coefficient extraction we obtain
$$P(z) = \sum_{p=0}^k \frac{(-1)^{k-p}}{(k-p)!} 
\frac{1}{1-pz} \frac{1}{p!}
= \frac{1}{k!} \sum_{p=0}^k {k\choose p} \frac{(-1)^{k-p}}{1-pz}.$$
Now to see that this is equal to 
$$Q(z) = \prod_{p=1}^k \frac{z}{1-pz}$$
we can use partial fractions by residues for rational functions,
getting
$$\mathrm{Res}_{z=1/m} Q(z)
= \mathrm{Res}_{z=1/m} \frac{z/m}{1/m-z}
\prod_{p=1}^{m-1} \frac{z}{1-pz}
\prod_{p=m+1}^k \frac{z}{1-pz}
\\ = -\frac{1}{m^2}
\prod_{p=1}^{m-1} \frac{1/m}{1-p/m}
\prod_{p=m+1}^k \frac{1/m}{1-p/m}
\\ = -\frac{1}{m^2}
\prod_{p=1}^{m-1} \frac{1}{m-p}
\prod_{p=m+1}^k \frac{1}{m-p}
= -\frac{1}{m^2} \frac{1}{(m-1)!} \frac{(-1)^{k-m}}{(k-m)!}
\\ = -\frac{1}{m} \frac{1}{m!} \frac{(-1)^{k-m}}{(k-m)!}.$$
On the other hand
$$\mathrm{Res}_{z=1/m} P(z)
= \mathrm{Res}_{z=1/m} 
\frac{1}{k!} {k\choose m} \frac{(-1)^{k-m}}{1-mz}
\\ = - \frac{1}{m} \mathrm{Res}_{z=1/m} 
\frac{1}{k!} {k\choose m} \frac{(-1)^{k-m}}{z-1/m}
= -\frac{1}{m} \frac{1}{m!} \frac{(-1)^{k-m}}{(k-m)!}.$$
The residues  are equal, all the  poles are simple,  the two functions
are rational  and have  no other poles,  hence we have  equality. This
concludes the proof.
