Combinatorics: set partitions of a poset 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.
 A: 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.
