[proofplan] Assume, for contradiction, that $\omega=d\lambda$ is symplectic. The symplectic condition forces $\dim M=2n$ with $n\ge 1$, and the top exterior power $\omega^n/n!$ gives the symplectic orientation and a positive volume form. On the other hand, since $\omega$ is exact and closed, the graded Leibniz rule gives $\omega^n=d(\lambda\wedge\omega^{n-1})$. [Stokes' theorem](/theorems/1530) on the compact boundaryless manifold then makes the integral of $\omega^n$ vanish, contradicting positivity of the symplectic volume. [/proofplan] [step:Reduce to the even-dimensional case forced by a symplectic form] Suppose, toward a contradiction, that there exists $\lambda\in\Omega^1(M)$ such that \begin{align*} \omega:=d\lambda \end{align*} is a symplectic form on $M$. Since a symplectic form is nondegenerate on each tangent space, each tangent space $T_xM$ has even dimension. Thus there is an integer $n\ge 1$ such that \begin{align*} \dim M=2n. \end{align*} The inequality $n\ge 1$ follows from the hypothesis $\dim M>0$. [/step] [step:Write the top exterior power as an exact form] Since $\omega=d\lambda$, the [nilpotence of the exterior derivative](/theorems/3913) gives \begin{align*} d\omega=d(d\lambda)=0. \end{align*} Define the $(2n-1)$-form \begin{align*} \alpha:=\lambda\wedge\omega^{n-1}\in\Omega^{2n-1}(M). \end{align*} Using the graded Leibniz rule for the [exterior derivative](/theorems/1525) and the fact that $\deg\lambda=1$, we obtain \begin{align*} d\alpha=d\lambda\wedge\omega^{n-1}-\lambda\wedge d(\omega^{n-1}). \end{align*} Because $d\omega=0$, the exterior derivative of $\omega^{n-1}$ is zero. Therefore \begin{align*} d\alpha=\omega\wedge\omega^{n-1}=\omega^n. \end{align*} [guided] The goal is to show that the top-degree form $\omega^n$ is exact, because exact top-degree forms integrate to zero on compact manifolds without boundary by Stokes' theorem. Since $\omega=d\lambda$, the natural primitive candidate is obtained by removing one factor of $\omega$ and replacing it by $\lambda$. Define \begin{align*} \alpha:=\lambda\wedge\omega^{n-1}\in\Omega^{2n-1}(M). \end{align*} This form has degree $1+2(n-1)=2n-1$, so $d\alpha$ has degree $2n$, the same degree as $\omega^n$. We now compute $d\alpha$ using the graded Leibniz rule. Since $\lambda$ has degree $1$, the sign in the second term is negative: \begin{align*} d(\lambda\wedge\omega^{n-1})=d\lambda\wedge\omega^{n-1}-\lambda\wedge d(\omega^{n-1}). \end{align*} The first term is exactly \begin{align*} d\lambda\wedge\omega^{n-1}=\omega\wedge\omega^{n-1}=\omega^n. \end{align*} For the second term, we use $d\omega=d(d\lambda)=0$. Since $\omega^{n-1}$ is a wedge product of copies of the closed $2$-form $\omega$, the graded Leibniz rule gives $d(\omega^{n-1})=0$. Hence the second term vanishes, and therefore \begin{align*} d\alpha=\omega^n. \end{align*} So the top exterior power of the supposed symplectic form is exact. [/guided] [/step] [step:Apply Stokes theorem to force the total integral to vanish] Equip $M$ with the symplectic orientation determined by the volume form $\omega^n/n!$, whose existence follows from [citetheorem:10035]. Since $M$ is compact and has no boundary, Stokes' theorem for compact manifolds without boundary gives \begin{align*} \int_M d\alpha=0. \end{align*} Using $d\alpha=\omega^n$, this becomes \begin{align*} \int_M \omega^n=0. \end{align*} [/step] [step:Use positivity of the symplectic volume to obtain a contradiction] By [citetheorem:10035], the form \begin{align*} \Omega:=\frac{\omega^n}{n!} \end{align*} is a nowhere-vanishing positive volume form for the symplectic orientation. Since $M$ is nonempty and compact, the integral of this positive volume form is strictly positive: \begin{align*} \int_M \Omega>0. \end{align*} Multiplying by $n!>0$ gives \begin{align*} \int_M \omega^n=n!\int_M\Omega>0. \end{align*} This contradicts the equality $\int_M\omega^n=0$ obtained from Stokes' theorem. Hence no $\lambda\in\Omega^1(M)$ can have $d\lambda$ symplectic. [/step]