The Khovanov cube differential squares to zero

Fix a state $v$ and two distinct coordinates $i<j$ with $v(c_i)=v(c_j)=0$. The two length-two paths from $v$ to $v+e_i+e_j$ induce equal unsigned maps: saddles on disjoint circles commute as tensor maps; when the two saddles meet one component, the three possible merge/split patterns give respectively associativity, coassociativity, or one of the two equal Frobenius composites in [F1]. These are all possibilities because each edge changes one smoothing by one saddle.

The sign on the path that changes $i$ then $j$ has exponent ${\sum }_{r<i}v(c_r)+{\sum }_{r<j}v(c_r)+1$, while the path that changes $j$ then $i$ has exponent ${\sum }_{r<j}v(c_r)+{\sum }_{r<i}v(c_r)$. Their signs are opposite, so the two equal unsigned composites from step 1.1 cancel in the sum defining the component of $d^2$ from $v$ to $v+e_i+e_j$. If there are fewer than two crossings, no length-two paths occur and $d^2$ is the empty sum; otherwise every term of $d^2$ lies in exactly one such pair. Hence $d^2=0$.