This is a long comment. To emphasize that you must interpret the hypothesis as being that $\omega^n (p) \ne 0$ for all $p\in M$, let me give a counterexample with the other interpretation (@user10354138). Let $M=S^4\times S^4$, with obvious projection maps $\pi_i$, $i=1,2$, to the $i$th factor. Choose any nonzero exact $2$-form $\eta$ on $S^4$ with the property that $\eta^2 = \eta\wedge\eta$ is not identically $0$. For example, take a compactly-supported exact $2$-form on $\Bbb R^4$, which extends naturally to be a form on $S^4$. [To be explicit, take $\rho_1$ to be a smooth function that is $1$ on the unit ball in $\Bbb R^3$ and $0$ outside the ball of radius $2$, and let $\rho_2$ be $1$ on the ball of radius $1$ centered at $(1,0,0,0)$ and $0$ outside the ball of radius $2$ centered at that point. Let $\eta = d(\rho_1\,dx-\rho_2\,dy)$. Note that $\eta^2 = d\rho_1\wedge d\rho_2\wedge dx\wedge dy$ will be nonzero at certain points of the intersection of the two balls of radius $2$.] Now let $\omega = \pi_1^*\eta + \pi_2^*\eta$. Then $\omega^4 = 6\pi_1^*(\eta^2)\wedge\pi_2^*(\eta^2)$ will be not identically zero on $S^4\times S^4$, and yet $H^2(S^4\times S^4) = 0$.