By way of enrichment I would like to point out that this can also be
done by Power Group Enumeration as defined by Harary and
Palmer. Here we have that the slot permutation group is the identity
$E_n$ with cycle index $Z(E_n) = a_1^n$ and the object permutation
group is the symmetric group $S_n$ with cycle index given by the
exponential formula
$$Z(S_n) = [z^n] \exp\left(\sum_{l\ge 1} a_l \frac{z^l}{l}\right).$$
Now to apply PGE we must cover the single permutation $a_1^n$ with
cycles from the permutations in $Z(S_n).$ Clearly if the latter has
$k$ fixed points this can be done in $k^n$ ways. Hence we introduce
the mixed generating function with fixed points marked
$$G(z, u) = \exp\left(uz-z+\log\frac{1}{1-z}\right).$$
The answer is thus given by (we turn $u^k z^n$ into $k^n z^n$)
$$ [z^n] \left. \left(u\frac{\partial}{\partial u}\right)^n G(z, u)
\right|_{u=1}
\\ = [z^n] \left. \left(u\frac{\partial}{\partial u}\right)^n
\frac{1}{1-z} \exp(z(u-1))
\right|_{u=1}
\\ = \left. \left(u\frac{\partial}{\partial u}\right)^n
\sum_{q=0}^n \frac{(u-1)^q}{q!}
\right|_{u=1}
\\ = \left. \left(u\frac{\partial}{\partial u}\right)^n
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} u^p
\right|_{u=1}
\\ = \sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} p^n.$$
We recognize Stirling numbers at this point but we may also continue
with the EGF
$$F(w) = \sum_{n\ge 0} \frac{w^n}{n!}
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} p^n
\\ = \sum_{n\ge 0} \frac{w^n}{n!}
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} n! [v^n] \exp(pv)
\\ = \sum_{n\ge 0} w^n [v^n]
\sum_{q=0}^n \frac{1}{q!}
\sum_{p=0}^q {q\choose p} (-1)^{q-p} \exp(pv)
\\ = \sum_{n\ge 0} w^n [v^n]
\sum_{q=0}^n \frac{1}{q!}
(\exp(v)-1)^q.$$
Now since $\exp(v)-1=v+\cdots$ we may extend $q$ beyond $n$ as there
is no contribution to the coefficient extractor $[v^n]$ in that case,
getting
$$\sum_{n\ge 0} w^n [v^n]
\sum_{q\ge 0} \frac{1}{q!}
(\exp(v)-1)^q
\\ = \sum_{n\ge 0} w^n [v^n] \exp(\exp(v)-1)
= \exp(\exp(w)-1)$$
and we see that we indeed have Bell numbers here as observed in the
reply that was first to appear. Note also that PGE rests on the
Burnside lemma which was suggested in the comments.
