 Solution

Remove the common Grassmann parameter and write the BRST transformation as the odd derivation $s$:The last two equations immediately give $s^2{\bar{c}}^a=s^2{B}^a=0$. Applying $s$ to the ghost and using the graded product rule givesThe cancellation is the Jacobi identity for the $SU(2)$ structure constants together with anticommutation of the ghost fields. For the gauge field, the component calculation isThe graded product rule givesso the two terms cancel. Thus $s^2$ vanishes on every elementary field.

For two independent Grassmann parameters, ${\delta }_1={\eta }_{1}s$ and ${\delta }_2={\eta }_{2}s$ satisfy ${\delta }_1{\delta }_{2}O={\eta }_1{\eta }_2{s}^{2}O=0$. Since $s$ obeys the graded Leibniz rule, induction extends $s^{2}O=0$ from the generators $A,c,\bar{c},B$ to every polynomial $O(A,c,\bar{c},B)$. Hence the BRST transformations are nilpotent.