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 $n \,|\, n-1\,|\, \cdots \,|\,2 \,|\, 1\,$ .