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.

I have tried using the recursion formulas of counting doubly surjective functions which didn't help much. My thought was the alternate signs appear for some Moebius maps on some partial order on the disjoint set of surjective maps from $n+j$ points to $j$ points up to permuting the image points (which explains division by $j!$). My definition of reshuffling images which give a partial order which give $0$ to distant points.