A decomposable differential form is one that can be written as a product of 1-forms. This is a non-trivial assertion; linear combinations are not allowed.

It is easy to see geometrically that all 2-forms in $\RR^3$ must be decomposable. In three dimensions, a 2-form can be thought of as the normal vector to a plane (more generally, a surface), and the wedge product mimics the cross product. But any vector in $\RR^3$ can be written as the cross product of two other vectors: choose any two linearly independent vectors in the plane orthogonal to the given vector, and rescale appropriately.

To find an example of a 2-form that it not decomposable, we need four dimensions, say with coordinates $w,x,y,z$. Then \begin{equation} \alpha = dx\wedge dy + dz\wedge dw \end{equation} is not decomposable. Why not? If $\alpha$ were decomposable, then we could write $\alpha=\beta\wedge\gamma$, for some 1-forms $\beta$ and $\gamma$. Clearly $\beta\wedge\gamma\wedge\beta\wedge\gamma=0$, but direct computation shows that $\alpha\wedge\alpha\ne0$.

We have in fact proved that $\alpha\wedge\alpha=0$ for any decomposable form, although the converse is false. Can you find an example of a differential form that is not decomposable, yet which “squares” to zero using the wedge product?

We are now able to provide an elegant argument that ($n-1$)-forms in $n$ dimensions are decomposable. Let $\alpha\in\bigwedge^{n-1}$. Then $*\alpha\in\bigwedge^1$ is a 1-form, and we can expand $*\alpha$ to a basis $\{\tau^1=*\alpha,\tau^2,…,\tau^n\}$ of $\bigwedge^1$. Since the $\tau^i$ are linearly independent, \begin{equation} \tau^1 \wedge … \wedge \tau^n = f\omega \end{equation} for some nonzero function $f$, which further implies that 1) \begin{equation} *\tau^1 = h\, \tau^2 \wedge … \wedge \tau^n \end{equation} for some nonzero function $h$. But now, since $**=\pm1$, \begin{equation} \alpha = \pm {*}(*\alpha) = \pm {*}\tau^1 = \pm h\, \tau^2 \wedge … \wedge \tau^n \end{equation} and we're done.