generating function of a sequence $a(m,k)$ i am interested if exist a explicit expresion of sequence given by
$$\sum_{k\geq 0}a(m,k)\frac{x^k}{k!}=\frac{1}{m!}(e^x-x-1)^m$$
I have tried to use binomial theorem but without any success
 A: We have that
$$n! [z^n] \frac{1}{m!} (\exp(z)-z-1)^m
\\ = n! [z^n] \frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} z^{m-q} (\exp(z)-1)^q
\\ = n! \frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q!
[z^n] z^{m-q} \frac{(\exp(z)-1)^q}{q!}
\\ = n! \frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q!
 [z^{n-(m-q)}] \frac{(\exp(z)-1)^q}{q!}
\\ = n! \frac{1}{m!} 
\sum_{q=0}^m {m\choose q} (-1)^{m-q} q!
\frac{1}{(n-(m-q))!} {n-(m-q)\brace q}
\\ = \sum_{q=0}^m {n\choose m-q}  (-1)^{m-q}
{n-(m-q)\brace q}.$$
If desired we may write this as
$$\sum_{q=0}^m {n\choose q}  (-1)^{q}
{n-q\brace m-q}.$$
These numbers appeared at the following MSE link.
We have the species
$$\mathfrak{P}_{=m}(\mathfrak{P}_{\ge 2}(\mathcal{Z}))$$
so this equation says in fact
$${n\brace m}_{\ge 2} = \sum_{q=0}^m {n\choose q}  (-1)^{q}
{n-q\brace m-q}.$$
This is  inclusion-exclusion of course where we  remove set partitions
into  $m$  sets  containing  singletons.   We  first  choose  the  $q$
singletons and combine them with an arbitrary set partition into $m-q$
sets of the remaining elements.  The poset is ordered according to set
inclusion of the sets of singletons.
