In defining a cohomology theory on sheaves of projective spaces, we define $$\mathscr{S}_{[k]} = \prod_{\left| J \right| = k+1} \mathscr{S}_J.$$ So essentially, we take the product of the $\mathscr{S}_J$, where the cardinality of the non-empty subset $J \subset \{ 0, ..., n \}$ is precisely $k+1$. Note that the $\mathscr{S}_J$ are the rings produced by the sheaf of rings $\mathscr{S}$ on the projective space.
Moreover, in any cohomology theory we need boundary maps $\partial$. We consider that $\mathscr{S}_J \subset \mathscr{S}_{[k]}$ and $\mathscr{S}_{J'} \subset \mathscr{S}_{[k+1]}$, and the boundary maps will be a matrix homomorphism. For each entry we have $$\partial_k^{J J'} = \begin{cases} \epsilon \cdot r_{J'}^J, & J \subset J', \\ 0, & \text{otherwise}. \end{cases}$$
Where $\epsilon = (-1)^{\left| S_{J J'} \right|}$, and $S_{J J'} = \{ s \in J' \ \vert \ s < j \}$.
I want to verify that the composite $$\mathscr{S}_{[k]} \xrightarrow{ \ \ \partial_k \ \ } \mathscr{S}_{[k+1]} \xrightarrow{ \ \ \partial_{k+1} \ \ } \mathscr{S}_{[k+2]},$$ is zero as it should be for cohomology.
I've attempted a proof by cases, taking the cases when $J \subset J'$, when $J \not \subset J'$, when $\left| J J' \right|$ is odd or even; but the proof is not very nice.