I'm trying to prove the following: If $(M, \omega)$ is a symplectic manifold and $[\omega]=0$ then $[\omega^n]=0$, where $[\omega]$ is the De Rham cohomology class of $\omega$. Well what I've done was:
Since $\omega$ is exact $\omega=d\tau$ for some $1$-form. Since $\omega^n=d\tau\wedge \omega^{n-1}$ then $$d(\tau\wedge \omega^{n-1})=d\tau \wedge \omega^{n-1}+(-1)^n \tau \wedge d\omega^{n-1}=\omega^n+(-1)^n \tau \wedge d\omega^{n-1},$$ hence $\omega^n=(-1)^{n+1}\tau\wedge d\omega^{n-1}+d(\tau\wedge \omega^{n-1})$. So, for finishing the proof if suffices showing $(-1)^{n+1}\tau\wedge d\omega^{n-1}=0$ but I can't see how to do it.. Can anyone help me? Thanks..
The statement holds for all $n$ and all exact forms $\eta$, so let's use induction.
Since $[\eta] = 0$, we have that $\eta$ is exact and hence $\eta = d\tau$ for some $\tau$.
To see that $[\eta \wedge \eta] = 0$, note that $$\eta \wedge \eta = d\tau \wedge d\tau = d(\tau \wedge d\tau).$$
Now suppose that $[\eta^{n-1}] = 0$. Then $\eta^{n-1} = d\alpha$ for some $\alpha$, and we have that \begin{align*} \eta^n & = \eta \wedge \eta^{n-1} \\ & = \eta \wedge d\alpha \\ & = (-1)^{\deg(\eta)}d(\eta \wedge \alpha), \end{align*} where the last equality is true since $d\eta = 0$ by assumption.
Hence by induction if $[\eta] = 0$ for some differential form $\eta$, $[\eta^n] = 0$ for all $n$.