Let $g(n,k)$ be the number of chains 


$$ A\_k \subset A\_{k-1} \subset\dots\subset A\_1 \subset A\_0 $$


of $k$ *proper* subset inclusions, where $A\_k\neq\emptyset$ and $A\_0$ is a standard $n$-element set. Then


$$ \sum\_{k\ge 0} (-1)^k g(n,k) = (-1)^{n-1}. $$


I can prove this by a fairly boring and unilluminating induction. (And also by a category-theoretic argument involving traces of geometric realizations in derivators, which is how I first noticed it.) Does it have a nicer combinatorial proof, by (say) bijections, generating functions, Mobius inversion, etc.?



Counting chains of inclusions
A chain
$$ A\_k \subset A\_{k-1} \subset\dots\subset A\_1 \subset A\_0 $$
can be represented by the ordered partition $(B\_1, B\_2, \dots, B\_{k+1})$ of the set $A\_0=\{1, 2, \dots, n\}$ where $B\_1=A\_k$, $B\_2=A\_{k-1}-A\_k$, $\dots,$ $B\_{k+1}=A\_0-A\_1$. 


First, a generating function proof. If we wanted to count such chains (or ordered partitions) without signs, the exponential generating function would be $\sum\_{j=0}^\infty (e^x-1)^j = 1/(2-e^x)$ (<http://oeis.org/A000670>). With signs, the generating function is
$$\sum\_{j=0}^\infty (-1)^{j+1}(e^x-1)^j=
-e^{-x} = \sum\_{n=0}^\infty (-1)^{n+1} \frac{x^n}{n!}.$$


For a combinatorial proof using a sign reversing involution that changes the parity of the number of blocks, write the entries of each block in increasing order, with a bar between blocks, so $\{2,4,6\}\{1,3\}\{5\}$ would be written as $ 2\, 4\, 6 \,|\, 1\, 3 \,|\, 5 $. Find the first position, if there is one, where a number is followed by a larger number. If there is a bar there, remove it, and if there is no bar there then put one in. So our example would be mapped to $ 2 \,|\, 4 \,6 \,|\, 1 \,3 \,|\, 5 $ . As another example, $4 \,|\, 3 \,|\, 1 \,2\,|\,$ would map to $4 \,|\, 3\,|\, 1 \,|\, 2$ . The only ordered partitions not paired up are of the form $n \,|\, n-1\,|\, \cdots \,|\,2 \,|\, 1\,$ .


