Let $M$ be a compact manifold of dimension $2n$. Let $\omega$ be a 2 form on $M$ such that the induced bundle map $\tilde{\omega}: TM \to T^*M$ defined by $\tilde{\omega}(X)(Y) = \omega(X,Y)$ is a bundle isomorphism. Show that if $d \omega = 0$, then $[\omega^n] \in H_{dR}^{2n}(M)$ is nonzero.
I recognize that on oriented compact manifolds top cohomology is nontrivial, but this question doesn't seem to assume that. I tried assuming $\omega^n$ was exact and getting some contradiction but didn't get anywhere. Any advice would be appreciated.
Or Eisenberg's comment goes in the correct direction, but it doesn't appear to quite work, since as written $\omega^n$ cannot possibly be $\det(a_{ij})$ times the volume form, since $\omega^n$ has degree $n$ in the entries of the matrix, but $(a_{ij})$ appears to be $2n\times 2n$.
Instead, let's do the following. Work on a framed open set $U$. Choose $X_1\in TM(U)$, choose $X_2$ such that $\omega(X_1,X_2)=1$ (since $\omega$ is nondegenerate). Then $\ker \omega(X_1,-)\cap \ker \omega(X_2,-)$ is $2n-2$ dimensional, so inductively choose $X_{2i-1}$, $X_{2i}$ such that $\omega(X_{2i-1},X_{2i})=1$ and when $i<j$, $\omega(X_{i},X_{j})=0$ otherwise. We may have to reduce the size of the open subset, so that we can select all of our vector fields.
Then given our basis $X_1,\ldots,x_{2n}$, let $\theta_1,\ldots,\theta_{2n}$ be the dual basis for $TM^*$. Observe that by construction of our basis, $$\omega = \sum_{i=1}^n\theta_{2i-1}\wedge \theta_{2i}.$$ Thus $\omega^n = n! \theta_1\wedge \cdots \wedge \theta_{2n}$. The $n!$ comes from the $n!$ ways to permute the $n$ 2-forms that make up $\omega$, since each permutation arises exactly once as a term in the product. All other terms in the product contain repeated 1-forms, and thus vanish.
Thus locally $\omega^n$ is a nowhere vanishing $2n$-form, so globally $\omega^n$ is a nowhere vanishing $2n$-form, i.e., a volume form.
On any (connected) compact manifold a volume form has nonzero integral, see here for a justification/explanation. Thus $\omega^n$ has nonzero integral, so its cohomology class is nonzero.
You may notice that we haven't used that $d\omega=0$. This is fine. Wikipedia confirms that $\omega^n$ is a volume form on almost symplectic manifolds as well.