Let $\pi\_n$ be the poset of all set partitions of $\{1,...,n\}$ ordered by refinement, $\sigma = \{B\_1,...,B\_k\}$ be a set partition with blocks $B\_i$, and $\max(B\_i)$ be the maximum value in the block $B\_i$. I'm trying to prove the following:


$$\Sigma\_{\sigma\in\pi\_{n-1}}\Pi\_{B\_i\in\sigma}(-1)^{|B\_i|-1}(|B\_i|-1)!(n+\max(B\_i))=(2n-1)!!$$


I've already coded this in Mathematica to check that it's true until n=11 (after which the numbers become too large):



```
<<Combinatorica`

f[x_]:=(-1)^(x-1)*(x-1) !

Total[Times @@@ Map[f, Map[Length, SetPartitions[n-1], {2}], {2}]*Times @@@ (Map[Max, SetPartitions[n-1], {2}] + n) ] == Factorial2[2*n-1]

```

Any thoughts would be greatly appreciated.



Combinatorics: set partitions of a poset
Let us prove even more general formula that Douglas Zare's one from the comments. Let $a\_1>\dots>a\_n$ be real numbers. Then
$$
 \sum\_{\sigma\in\pi\_{n}}
 \prod\_{B\_i\in\sigma}(-1)^{|B\_i|-1}(|B\_i|-1)!a\_{\min B\_i}
 =\prod\_{i=1}^n(a\_i-i+1).
$$
(The required formula follows by setting $a\_i=2(n+1)-i$ and shifting $n$ by $1$.)


To see this, rewrite the right-hand side as
$$
 \prod\_{i=1}^n\left(a\_i-\sum\_{1\leq j<i}u\_{i,j}\right),
$$
where $u\_{i,j}=1$ for $j<i$, and expand all the brackets. We get the sum of the terms each having the form
$$
 (-1)^{n-|I|}\prod\_{i\in I}a\_i\prod\_{i\notin I}u\_{i,j(i)},
$$
where $I$ is a subset of $\{1,\dots,n\}$ containing $1$. To each such term, put into correspondence a partition into $|I|$ parts, where $a\_i$ ($i\in I$) are the maximal elements of the parts, and each $u\_{i,j}$ is interpreted as `$a\_i$ and $a\_j$ lie in the same set of partition'. 


This way, each partition $(B\_i)$ will correspond to exactly $\prod\_i (|B\_i|-1)!$ terms, all having the sign we need. Thus, after substituting $u\_{ij}=1$ we obtain the required formula.


