How do I prove :
$\sum_{j=2}^{n} (-1)^j {\frac {M(n+j,j;2)}{j!}} = (-1)^n n! + 1$?
where $M(n+j,j;2)$ is the multinomial sum $M(n+j,j;2) = \sum_{t_1 + t_2 + \dotsc + t_j = n+j, t_k \geq 2} {n+j \choose t_1 \dotsc t_k}$ which denotes the number of surjective functions from $n+j$ points to $j$ points with at least two elements in each pre-image.
$M(n+j,j;2)$ is the number of ways of putting $n+j$ distinct balls into $j$ distinct urns such that each urn contains at least two balls. This is also the number of functions from a domain of $n+j$ objects (the balls) into a range of $j$ objects (the urns) which are doubly-surjective, i.e. every element of the range has at least two pre-images.
This latter problem has been solved in Notes on Doubly-Surjective Finite Functions by Walsh, who shows that $$M(n+j,j;2) = \displaystyle\sum_{k=0}^j\left[(-1)^k{j \choose k}\sum_{i=0}^k{k \choose i}\frac{(n+j)!}{(n+j-i)!}(j-k)^{n+j-i}\right]$$