We can of course multiply any differential forms together, not just 1-forms. So let $\alpha\in\bigwedge^p$ and $\beta\in\bigwedge^q$. Clearly we must have $\alpha\wedge\beta\in\bigwedge^{p+q}$. For example, if \begin{align} \alpha &= dx\wedge dy \\ \beta &= dz\wedge du\wedge dv \end{align} then \begin{equation} \alpha\wedge\beta = dx\wedge dy\wedge dz\wedge du\wedge dv \end{equation}

What about $\beta\wedge\alpha$? Direct computation shows that, in this case, $\beta\wedge\alpha=\alpha\wedge\beta$. Why? Because you have to move three 1-forms through two 1-forms, for a total of six sign changes. Thus, in general, \begin{equation} \beta\wedge\alpha = (-1)^{pq} \> \alpha\wedge\beta \end{equation} Our original antisymmetry rule for 1-forms is just the special case $p=q=1$. The simple argument given in §1.4 shows that \begin{equation} \alpha\wedge\alpha = 0 \end{equation} for any $p$-form with $p$ odd, but this argument fails for $p$ even. Can you find a $p$-form $\alpha$ such that $\alpha\wedge\alpha\ne0$?