[proofplan] The assertion is pointwise, so we fix a point $p\in M$ and compute in the symplectic [vector space](/page/Vector%20Space) $(T_pM,\omega_p)$. A symplectic basis puts $\omega_p$ into standard form, and the $n$-fold wedge expansion then reduces to exactly $n!$ copies of the ordered basis volume element. This proves nonvanishing at every point. Since $\omega$ is smooth, the resulting top form is smooth and nowhere vanishing, hence determines an orientation. [/proofplan]

[step:Compute the top wedge power in a symplectic basis]Fix $p\in M$. If $n=0$, then $\omega^0/0!$ is the constant $0$-form $1$ on $M$, so it is nowhere vanishing. Assume now that $n\ge 1$. The tangent space $T_pM$ is a $2n$-dimensional symplectic vector space with symplectic form $\omega_p$. By the [[Symplectic Basis Theorem](/theorems/10037)][citetheorem:10037], there is a basis \begin{align*} (e_1,\dots,e_n,f_1,\dots,f_n) \end{align*} of $T_pM$ such that \begin{align*} \omega_p(e_i,e_j)=0 \end{align*} for all $1\le i,j\le n$, \begin{align*} \omega_p(f_i,f_j)=0 \end{align*} for all $1\le i,j\le n$, and \begin{align*} \omega_p(e_i,f_j)=\delta_{ij} \end{align*} for all $1\le i,j\le n$. Let \begin{align*} (e_1^*,\dots,e_n^*,f_1^*,\dots,f_n^*) \end{align*} be the [dual basis](/theorems/414) of $T_p^*M$. With this convention, \begin{align*} \omega_p=\sum_{i=1}^n e_i^*\wedge f_i^*. \end{align*} Define the $2$-forms $\alpha_i\in\Lambda^2T_p^*M$ by \begin{align*} \alpha_i=e_i^*\wedge f_i^*. \end{align*} Then $\omega_p=\sum_{i=1}^n\alpha_i$. Since each $\alpha_i$ has degree $2$, the wedge product is commutative on the $\alpha_i$: \begin{align*} \alpha_i\wedge\alpha_j=\alpha_j\wedge\alpha_i. \end{align*} Also $\alpha_i\wedge\alpha_i=0$, because $e_i^*$ and $f_i^*$ each appear twice. Therefore, in the expansion of $(\alpha_1+\cdots+\alpha_n)^n$, every term with a repeated index vanishes, and every nonzero term contains each of $\alpha_1,\dots,\alpha_n$ exactly once. There are $n!$ such terms, indexed by permutations of $\{1,\dots,n\}$, and all have the same sign because the $\alpha_i$ commute. Hence \begin{align*} \omega_p^n=n!\,\alpha_1\wedge\cdots\wedge\alpha_n. \end{align*} Substituting the definition of $\alpha_i$ gives \begin{align*} \frac{1}{n!}\omega_p^n=e_1^*\wedge f_1^*\wedge\cdots\wedge e_n^*\wedge f_n^*. \end{align*} The final wedge product is the exterior product of the dual basis covectors, so it is a nonzero element of $\Lambda^{2n}T_p^*M$. Thus \begin{align*} \left(\frac{1}{n!}\omega^n\right)_p\ne 0. \end{align*}[/step]

[guided]We prove nonvanishing at a single point $p\in M$, because a differential form is nowhere vanishing precisely when its value at every point is nonzero. If $n=0$, then $M$ has dimension $0$ and the convention for exterior powers gives $\omega^0/0!=1$, the constant $0$-form, which is nonzero at every point. We now assume $n\ge 1$. The vector space $T_pM$ has dimension $2n$, and $\omega_p$ is a nondegenerate alternating [bilinear form](/page/Bilinear%20Form) on it. By the [Symplectic Basis Theorem][citetheorem:10037], we may choose a basis \begin{align*} (e_1,\dots,e_n,f_1,\dots,f_n) \end{align*} of $T_pM$ satisfying the symplectic basis relations \begin{align*} \omega_p(e_i,e_j)=0 \end{align*} for all $1\le i,j\le n$, \begin{align*} \omega_p(f_i,f_j)=0 \end{align*} for all $1\le i,j\le n$, and \begin{align*} \omega_p(e_i,f_j)=\delta_{ij} \end{align*} for all $1\le i,j\le n$. Let \begin{align*} (e_1^*,\dots,e_n^*,f_1^*,\dots,f_n^*) \end{align*} denote the dual basis of $T_p^*M$. In this dual basis, the defining pairings above say exactly that \begin{align*} \omega_p=\sum_{i=1}^n e_i^*\wedge f_i^*. \end{align*} The reason this normal form is useful is that the top exterior power can now be computed by pure algebra. For each $1\le i\le n$, define \begin{align*} \alpha_i=e_i^*\wedge f_i^*\in\Lambda^2T_p^*M. \end{align*} Then $\omega_p=\alpha_1+\cdots+\alpha_n$. Since every $\alpha_i$ has degree $2$, swapping two such forms gives the sign $(-1)^{2\cdot 2}=1$, so \begin{align*} \alpha_i\wedge\alpha_j=\alpha_j\wedge\alpha_i. \end{align*} Also, \begin{align*} \alpha_i\wedge\alpha_i=(e_i^*\wedge f_i^*)\wedge(e_i^*\wedge f_i^*)=0, \end{align*} because an alternating wedge product vanishes whenever a covector occurs twice. Now expand \begin{align*} \omega_p^n=(\alpha_1+\cdots+\alpha_n)^n. \end{align*} Any term in this expansion with a repeated index contains $\alpha_i\wedge\alpha_i$ for some $i$, after commuting the degree-$2$ factors, and therefore vanishes. The only nonzero terms are those in which every one of $\alpha_1,\dots,\alpha_n$ appears exactly once. These terms are indexed by the $n!$ permutations of $\{1,\dots,n\}$. Since the $\alpha_i$ commute, each of those $n!$ terms equals $\alpha_1\wedge\cdots\wedge\alpha_n$. Therefore \begin{align*} \omega_p^n=n!\,\alpha_1\wedge\cdots\wedge\alpha_n. \end{align*} Dividing by $n!$ gives \begin{align*} \frac{1}{n!}\omega_p^n=\alpha_1\wedge\cdots\wedge\alpha_n. \end{align*} Using the definition of the $\alpha_i$, this becomes \begin{align*} \frac{1}{n!}\omega_p^n=e_1^*\wedge f_1^*\wedge\cdots\wedge e_n^*\wedge f_n^*. \end{align*} This is the exterior product of all covectors in a basis of $T_p^*M$, so it is a nonzero element of the one-dimensional vector space $\Lambda^{2n}T_p^*M$. Hence \begin{align*} \left(\frac{1}{n!}\omega^n\right)_p\ne 0. \end{align*}[/guided]

[step:Conclude that the pointwise nonzero top form determines an orientation] The point $p\in M$ was arbitrary, so the previous step proves that \begin{align*} \frac{1}{n!}\omega^n \end{align*} is nonzero at every point of $M$. Since $\omega\in\Omega^2(M)$ is smooth and the wedge product of smooth differential forms is smooth, the form $\omega^n/n!$ belongs to $\Omega^{2n}(M)$. A smooth nowhere-vanishing top-degree form is an orientation form: at each $p\in M$, an ordered basis $(v_1,\dots,v_{2n})$ of $T_pM$ is declared positive precisely when \begin{align*} \left(\frac{1}{n!}\omega^n\right)_p(v_1,\dots,v_{2n})>0. \end{align*} Because the form is smooth and nowhere zero, this pointwise choice varies smoothly in local frames and defines an orientation on $M$. Therefore $M$ is orientable, and the canonical orientation is the one represented by $\omega^n/n!$. [/step]