 Solution

Use the standard convention that the compact manifold is nonempty and has no boundary. Orient it by ${\omega }^m$. Since $d\omega =0$, every power ${\omega }^k$ is closed. If ${\omega }^k=d\eta$ for some $1\le k\le m$, thenagain by Leibniz and closedness. Stokes theorem would give ${\int }_M{\omega }^m=0$. But this integral is strictly positive in the symplectic orientation, a contradiction. ThereforeThe boundaryless hypothesis matters: a compact disk has a symplectic area form but no nonzero ordinary second cohomology class.