= Solution
Factor the odd parameter on the left, $\delta=\epsilon s$. The resulting <left-acting BRST differential> obeys the <graded Leibniz rule>
$$
s(XY)=(sX)Y+(-1)^{|X|}X(sY).
$$
For the odd <Grassmann field> $c$, the bracket in the transformation is a <graded commutator>: $[c,c]_{\rm gr}=2c^2$, not the identically zero ordinary commutator of a matrix with itself. Thus $sc=ic^2$, while $sA_\mu=D_\mu c$, $s\bar c=h$ and $sh=0$.
On the ghost,
$$
s^2c=i[(sc)c-c(sc)]=i[ic^2c-ic c^2]=0.
$$
On the gauge field, variation of the connection and the <adjoint covariant derivative> gives
$$
s^2A_\mu=D_\mu(sc)-i[sA_\mu,c]_{\rm gr}
=iD_\mu(c^2)-i[(D_\mu c)c+c(D_\mu c)]=0.
$$
Here $D_\mu$ is even and therefore obeys the ordinary product rule. Also $s^2\bar c=sh=0$ and $s^2h=0$, without using any field equation; this is off-shell nilpotence supplied by the <Nakanishi-Lautrup field>.
Applying the <graded Leibniz rule> twice cancels the two cross terms:
$$
s^2(XY)=(s^2X)Y+Xs^2Y.
$$
The square is consequently an even <graded derivation>. Since it vanishes on every generator, it vanishes inductively on every polynomial in the fields. Hence \b[$s^2\mathcal O=0$ for every such operator]. This genuine result is stronger than the automatic vanishing obtained by merely setting $\epsilon^2=0$; two independent transformation parameters also give a vanishing commutator.
Back to article page
