[proofplan] We extract two topological consequences from the single algebraic fact that the top power $\omega^n$ is a nowhere-vanishing $2n$-form on $M^{2n}$. Non-vanishing makes $\omega^n$ a volume form, which directly orients $M$ and, by [Stokes' Theorem](/theorems/???) on a closed manifold, prevents $\omega^n$ from being exact. Hence $[\omega^n] = [\omega]^n \neq 0$ in de Rham cohomology. The ring structure of $H^*_{\mathrm{dR}}(M)$ then forces $[\omega]^i \neq 0$ in every intermediate even degree $2i$. [/proofplan] [step:Identify $\omega^n$ as a nowhere-vanishing top-degree form] Define the $n$-th exterior power \begin{align*} \omega^n : M &\to \Lambda^{2n} T^*M \\ p &\mapsto \underbrace{\omega_p \wedge \cdots \wedge \omega_p}_{n \text{ times}}. \end{align*} By the [Non-degeneracy Characterisation of $\omega^n$](/theorems/???), the non-degeneracy of the symplectic form $\omega$ is equivalent to $\omega^n_p \neq 0$ for all $p \in M$. Since $\dim_{\mathbb R} M = 2n$, the bundle $\Lambda^{2n} T^*M$ is a real line bundle, so $\omega^n$ is a nowhere-vanishing global section of a line bundle, i.e.\ a volume form on $M$. [/step] [step:Orient $M$ via the volume form $\omega^n$] A volume form on a smooth manifold determines an orientation: declare an ordered frame $(v_1, \dots, v_{2n})$ of $T_pM$ to be positively oriented precisely when $\omega^n_p(v_1, \dots, v_{2n}) > 0$. This choice is continuous in $p$ because $\omega^n$ is a smooth section and never vanishes. This establishes part (1). [guided] We want to orient $M$, which means choosing a continuous choice of "positively oriented" ordered basis in each tangent space $T_pM$. Such a continuous choice exists iff $M$ admits a nowhere-vanishing section of $\Lambda^{2n} T^*M$: a volume form. We have just produced such a section, namely $\omega^n$. Concretely, for each $p \in M$ declare $(v_1, \dots, v_{2n})$ to be positively oriented at $p$ iff $\omega^n_p(v_1, \dots, v_{2n}) > 0$. This is well-defined because $\omega^n_p \neq 0$ and because a reordering of the frame multiplies the value of $\omega^n_p$ by the sign of the permutation, flipping the class. Continuity in $p$ holds because $\omega^n$ is smooth and nowhere zero, so the sign $\omega^n_p(v_1, \dots, v_{2n})$ cannot change when we vary $p$ and the $v_i$ smoothly. This is the orientation canonically attached to the symplectic structure. [/guided] [/step] [step:Show $[\omega^n] \neq 0$ in $H^{2n}_{\mathrm{dR}}(M)$ via Stokes] The form $\omega^n \in \Omega^{2n}(M)$ has degree equal to $\dim M$, so $d\omega^n \in \Omega^{2n+1}(M) = 0$; hence $\omega^n$ is closed and defines a class $[\omega^n] \in H^{2n}_{\mathrm{dR}}(M)$. Suppose towards contradiction that $[\omega^n] = 0$, i.e.\ $\omega^n = d\eta$ for some $\eta \in \Omega^{2n-1}(M)$. Since $M$ is closed (compact and without boundary) and oriented by part (1), [Stokes' Theorem](/theorems/???) applies and gives \begin{align*} \int_M \omega^n = \int_M d\eta = \int_{\partial M} \eta = 0. \end{align*} On the other hand, $\omega^n$ is the positively-oriented volume form from the previous step, so $\int_M \omega^n > 0$. This contradiction shows $[\omega^n] \neq 0$. [guided] The goal is to produce a non-zero de Rham cohomology class in the top degree $2n$. A natural candidate is $[\omega^n]$. We must verify two things: that it defines a class (closedness), and that the class is non-zero. Closedness is automatic: any $2n$-form on a $2n$-manifold is closed for dimensional reasons, since $\Omega^{2n+1}(M) = 0$. So $d\omega^n = 0$ and $[\omega^n] \in H^{2n}_{\mathrm{dR}}(M)$ is well-defined. To show $[\omega^n] \neq 0$, suppose for contradiction $\omega^n = d\eta$ for some $\eta \in \Omega^{2n-1}(M)$. We now invoke [Stokes' Theorem](/theorems/???), which requires (i) an oriented smooth manifold with boundary and (ii) a compactly supported smooth form of top degree. Our $M$ is closed by hypothesis, so compact without boundary; it is oriented by the previous step; and $\eta$ is smooth on a compact manifold, hence automatically compactly supported. Stokes gives \begin{align*} \int_M \omega^n = \int_M d\eta = \int_{\partial M} \eta = 0 \end{align*} because $\partial M = \varnothing$. But the orientation of the previous step was engineered precisely so that $\omega^n$ is a positively-oriented volume form, which means $\int_M \omega^n > 0$ (any volume form of the orientation has strictly positive total volume). This is a contradiction, so $[\omega^n] \neq 0$. [/guided] [/step] [step:Propagate non-vanishing to intermediate powers via the cohomology ring] De Rham cohomology carries a graded-commutative ring structure induced by the wedge product: the cup product on cohomology satisfies $[\alpha] \smile [\beta] = [\alpha \wedge \beta]$ for closed forms $\alpha, \beta$. In particular, since $d\omega = 0$, \begin{align*} [\omega^n] = [\omega]^n \in H^{2n}_{\mathrm{dR}}(M). \end{align*} Fix $i$ with $0 \leq i \leq n$. If $[\omega]^i = 0 \in H^{2i}_{\mathrm{dR}}(M)$, then by multiplicativity \begin{align*} [\omega]^n = [\omega]^i \smile [\omega]^{n-i} = 0 \smile [\omega]^{n-i} = 0, \end{align*} contradicting the previous step. Therefore $[\omega]^i \neq 0$, which in particular shows $H^{2i}_{\mathrm{dR}}(M) \neq 0$. This establishes part (2) and completes the proof. [guided] We have $[\omega^n] \neq 0$ in top degree, and we want to upgrade this to $H^{2i}_{\mathrm{dR}}(M) \neq 0$ for every intermediate $0 \leq i \leq n$. The bridge is the ring structure on $H^*_{\mathrm{dR}}(M)$: the wedge product descends to cohomology as the cup product, so we may speak of $[\omega]^i$. First, identify $[\omega^n]$ with $[\omega]^n$. Since $\omega$ is closed ($d\omega = 0$ is part of the symplectic axioms), each factor represents its own cohomology class and the product of the classes equals the class of the wedge product: \begin{align*} [\omega]^n = \underbrace{[\omega] \smile \cdots \smile [\omega]}_{n} = [\underbrace{\omega \wedge \cdots \wedge \omega}_{n}] = [\omega^n]. \end{align*} Now fix $0 \leq i \leq n$. The boundary cases are easy: $i = 0$ gives $[\omega]^0 = 1 \in H^0_{\mathrm{dR}}(M) \cong \mathbb R$ (constant function $1$), which is non-zero on a non-empty manifold; $i = n$ is precisely the previous step. For $0 < i < n$, argue by contradiction. Suppose $[\omega]^i = 0 \in H^{2i}_{\mathrm{dR}}(M)$. Multiplying by $[\omega]^{n-i}$ (which is a well-defined element of $H^{2(n-i)}_{\mathrm{dR}}(M)$) and using that the cup product is bilinear, hence $0 \smile [\omega]^{n-i} = 0$, we get \begin{align*} [\omega]^n = [\omega]^i \smile [\omega]^{n-i} = 0, \end{align*} contradicting $[\omega^n] \neq 0$ from the previous step. Hence $[\omega]^i \neq 0$, and in particular $H^{2i}_{\mathrm{dR}}(M) \neq 0$ since it contains the non-zero class $[\omega]^i$. This is exactly the assertion of part (2). [/guided] [/step]