I was trying to prove that every non-degenerate $2$-form cant be exact in an even dimensional smooth connected compact oriented manifold. I tried the following: Given a $2$-form $\omega$ that is exact, then it can be written as: $\omega = d \alpha$. Then:
$$ d( \alpha \wedge (d \alpha )^{n-1} ) = d \alpha \wedge (d \alpha)^{n-1} - \alpha \wedge d(( d \alpha)^{n-1})$$
this eventually leads us to $\omega^n$ but Im having a few doubts regarding these terms. In a rubish way:
$$ d \alpha \wedge (d \alpha)^{n-1}= d \alpha \wedge \frac{(d \alpha)^{n}}{d \alpha} = (d \alpha)^n = \omega^n$$
and for the second term I know that $d(d\alpha)=0$ but when trying a specific example (in $\mathbb{R}^3$):
$$ \alpha = xyz dx \longrightarrow d \alpha =\sum_{\mu=1}^3 \partial_{\mu}(xy)dx^{\mu} \wedge dx=xz dy \wedge dx + xy dz \wedge dx$$
$$ (d \alpha )^{3-1}= (xz dy \wedge dx + xy dz \wedge dx )^{3-1} \longrightarrow d ((d \alpha )^{3-1}) = ? $$
the elaboration in these calculations is confusing me. Also, does this only work for linear forms (linear in the dependence of $x$,$y$,$z$)? Any other proof would also be welcome.
Very shortly, if $\omega = {\rm d}\eta$ for some $\eta \in \Omega^1(M)$ and $M^{2n}$ is compact, then $\omega^{\wedge n}$ is a volume form. So $$0 \neq \int_M \omega^{\wedge n} = \int_M {\rm d}\eta \wedge \omega^{\wedge (n-1)} = \int_M {\rm d}(\eta \wedge \omega^{\wedge(n-1)}) = \int_{\partial M} \eta \wedge \omega^{n-1} = \int_{\varnothing} \eta \wedge \omega^{\wedge(n-1)} = 0.$$Hence we have that if $(M^{2n},\omega)$ is a compact symplectic manifold, then $H^2_{\rm dR}(M) \neq 0$. A similar argument shows that $H^{2k}_{\rm dR}(M) \neq 0$ for $1 \leq k \leq n$.