Problem: Suppose $w$ is a closed $2$-form on a compact manifold $M$ without boundary, with $\dim(M) = 2n$ for some $n$. Denoting $w^k = w \wedge \dots \wedge w$ to be the wedge product of $w$ with itself for $k$ times, we also suppose $w^n$ is a volume form on $M$. Prove that $w^k$ is not exact, for $k = 1, 2, \dots, n$.
Attempt: I'm not sure how to approach the problem, but I suspect it has something to do with Stoke's theorem. For instance, if $w^n$ is exact, write $w^n = d \alpha,$ then $0 \neq \int_M w^n = \int_M d \alpha = 0$, but what about $w^k$ for $k \lt n$? Any help is appreciated!
Suppose that $\omega^k=du$, $\omega^n=\omega^k\wedge\omega^{n-k}=du\wedge \omega^{n-k}$.
$d(u\wedge\omega^{n-k})=du\wedge\omega^{n-k}+(-1)^{2k-1}u\wedge d\omega^{n-k}=du\wedge\omega^{n-k}$, since $d\omega=0$ recursively, $d\omega^p=0$. We deduce that $\int_M\omega^n=\int_Md(u\wedge\omega^{n-k})=0$ contradiction.