Let $\genfrac{\{}{\}}{0pt}{}{n}{k}$ the Stirling number of second kind, where $k$ is the number of parts in the partition.
If we take the identity that transforms the polynomial base $x^k$ into the base $x(x-1)\ldots(x-k+1)$ and set $x=-1,$ we obtain that $$\sum_{k=0}^{n} (-1)^k k! \genfrac{\{}{\}}{0pt}{}{n}{k} = (-1)^n.$$
My question is to get this formula only by bijective means, by relating partitions on odd parts with partitions on even parts. I have been unsuccessful.
We sum up $(-1)^k$ over all ordered partitions $B=(B_1,B_2,\ldots,B_k)$ of $[n]=\{1,2,\ldots,n\}$. Consider the following pairing of such partitions: if $1\in B_j$ and $|B_j|>1$, then $B$ is paired with $$(B_1,\ldots,B_{j-1},B_j\setminus \{1\},\{1\},B_{j+1},\ldots,B_{n}).$$ So, all partitions are paired (and in each pair the sizes of partitions differ by 1) except those for which $B_1=\{1\}$. For them proceed similarly taking $2$ out of its part, etc. Finally you pair all ordered partitions except $(\{1\},\{2\},\ldots,\{n\})$. Thus the result.
