I'm trying to understand why $B(n-1)$ also counts the number of partitions of $[n]$ where not two consecutive integers appear in the same block.
Now the bell number $B(n-1)$ counts the number of partitions of the $n-1$-set $[n-1]$. Suppose I take any partition $\pi$ of $[n-1]$. Now taking $i,i+1,\dots,j$ to be a maximal sequence of two or more consecutive integers in a block, I can remove alternating integers $j-1$, $j-3$, $j-5$,... and put them in a block with $n$. Doing so for all sequences of consecutive integers in blocks of $\pi$ will then give a partition of $[n]$ with no two consecutive numbers.
I think this gives a needed bijection of the two things, but if I'm given a partition of $[n]$ with no two consecutive integers in a block, how can I reconstruct the partition of $[n-1]$ to see that it is indeed a bijection?