To prove Dobiński's formula we can use combinatorial species and their generating functions.
There's a species (\mathrm{Part}), such that (\mathrm{Part}(n)) is the set of partitions of an (n)-element set into nonempty subsets. By definition its generating function is
[ \displaystyle{
|\mathrm{Part}|(x) =
\sum_{n \ge 0} \frac{|\mathrm{Part}(n)|}{n!} x^n =
\sum_{n \ge 0} \frac{B_n}{n!} x^n } ]
since we call the cardinality (|\mathrm{Part}(n)|) the (n)th Bell number.
To put a partition on a finite set amounts to chopping it into a finite set of nonempty finite sets, so using a cool fact about species, we have
[ |\mathrm{Part}| = |\mathrm{Exp}| \circ |\mathrm{NE}| ]
where (\mathrm{Exp}) is the species 'being a finite set' and (\mathrm{NE}) is the species 'being a nonempty finite set'. There's one way for an (n)-element set to be a finite set so
[ |\mathrm{Exp}|(x) = \sum_{n \ge 0} \frac{x^n}{n!} = e^x ]
and similarly
[ |\mathrm{NE}|(x) = \sum_{n \ge 1} \frac{x^n}{n!} = e^x - 1 ]
Thus we have
[ |\mathrm{Part}|(x) = e^{e^x - 1} ]
and thus
[ \sum_{n \ge 0} \frac{B_n}{n!} x^n = e^{e^x - 1} ]
Now let's use this to prove Dobiński's formula! We start by calculating the right hand side another way. We have
[ \displaystyle{ e^{e^x} = \sum_{k \ge 0} \frac{e^{kx}}{k!}
= \sum_{k \ge 0} \frac{1}{k!} \sum_{n \ge 0} \frac{(kx)^n}{n!} }]
so
[ \displaystyle{ e^{e^x - 1} =
\frac{1}{e} \sum_{k \ge 0} \frac{1}{k!} \sum_{n \ge 0} \frac{(kx)^n}{n!}} ]
The coefficient of (x^n) in this power series must be (B_n/n!), so Dobiński's formula follows:
[ \displaystyle{ B_n = \frac{1}{e} \sum_{k = 0}^\infty \frac{k^n}{k!} } ]