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.
