Given the 1-form, $$\alpha = dz - pdx - qdy,$$ I can calculate $$d\alpha = -dp\wedge dx - dq \wedge dy = dx \wedge dp + dy \wedge dq, $$ but I'm not so sure how to calculate $d\alpha \wedge d\alpha$.
Could someone please point me in the right direction?
$$\begin{split} d\alpha\wedge d\alpha &=(dx\wedge dp+dy\wedge dq)\wedge (dx\wedge dp+dy\wedge dq)\\ &=dx\wedge dp\wedge dy\wedge dq+dy\wedge dq\wedge dx\wedge dp\\ &=2dx\wedge dp\wedge dy\wedge dq \end{split}$$
Use the fact that $\alpha\wedge \beta=(-1)^{\text{degree}(\alpha)\text{degree}(\beta)}\beta\wedge \alpha$ if $\beta$ and alpha are homogeneous.