If $\alpha $ is one form over some manifold $M$ $2n-1$ dimensional real, and $X= M\times (0,\infty)$. $r$ is the coordinate for the second factor. Define two form on $X$: $$\omega= d(r^2\alpha)$$ Then we have to calculate $\omega^n$. I am sorry if following doubt are too silly: My doubts are:
1- I think, $\omega^n:= \omega\wedge..\wedge\omega$, n times.
2- As $\omega$ is two form hence $\omega\wedge \omega \neq 0$. but for any one form $\alpha$, we must have $\alpha\wedge\alpha= 0$.[As $\alpha\wedge\alpha= c(\alpha\otimes \alpha- \alpha\otimes \alpha)$
3- What is the guarantee that $\omega^n\neq 0$, As in one form(as in 2nd part above), can we say when $\omega^n=0$ for any two form.