Prove that $d(d(w)) = 0$ for a differential $n$-form $w$, where $d$ is the exterior derivative

Question

I've managed to get that for the case of $w = a_Idx_I$, we have by the leibniz rule $d(d(w)) = d(d(a_I)) \wedge dx_I - da_I \wedge d(dx_I)$. The first term is $0$ because $a_I$ is a $0$-form. But why does the second term evaluate to $0$? How do I calculate $d(dx_I)$?