[proofplan] We use the bundle definition of forms of type $(p,q)$. The complex cotangent bundles $T^{*1,0}M$ and $T^{*0,1}M$ both have complex rank $n$, so their exterior powers vanish in degrees greater than $n$. Therefore, if $p>n$ or $q>n$, the vector bundle whose smooth sections define $\Omega^{p,q}(M)$ is the zero vector bundle, and its only smooth section is the zero section. [/proofplan] [step:Express $(p,q)$-forms as sections of exterior cotangent bundles] Let $E:=T^{*1,0}M$ denote the holomorphic cotangent bundle of $M$, and let $F:=T^{*0,1}M$ denote the antiholomorphic cotangent bundle of $M$. Since $M$ has complex dimension $n$, both $E$ and $F$ are complex vector bundles of rank $n$ over $M$. By definition, the space of smooth differential forms of type $(p,q)$ is $\Omega^{p,q}(M)=\Gamma\left(M,\Lambda^p E\otimes_{\mathbb C}\Lambda^q F\right)$, with the standard convention that exterior powers outside the allowed degree range are the zero bundle. [/step] [step:Show that exterior powers above rank vanish] Let $G\to M$ be a complex vector bundle of rank $n$. We claim that $\Lambda^k G$ is the zero vector bundle for every integer $k>n$. For each point $x\in M$, the fiber $G_x$ is a complex [vector space](/page/Vector%20Space) of dimension $n$. The fiber of $\Lambda^k G$ at $x$ is $\Lambda^k(G_x)$. If $k>n$, every alternating $k$-fold tensor on an $n$-dimensional vector space vanishes, because any list of $k$ vectors in $G_x$ is linearly dependent. Hence $\Lambda^k(G_x)=\{0\}$ for every $x\in M$. Therefore $\Lambda^k G$ is the zero vector bundle. [guided] The point is purely linear algebra in each fiber. A rank-$n$ complex vector bundle $G\to M$ assigns to each point $x\in M$ an $n$-dimensional complex vector space $G_x$. The exterior power bundle $\Lambda^k G$ is obtained fiberwise, so its fiber at $x$ is $(\Lambda^k G)_x=\Lambda^k(G_x)$. Now suppose $k>n$. In the vector space $G_x$, any $k$ vectors must be linearly dependent, since the dimension is only $n$. An alternating $k$-linear expression vanishes whenever its inputs are linearly dependent. Thus every decomposable element $v_1\wedge\cdots\wedge v_k$ in $\Lambda^k(G_x)$ is zero, and these decomposable elements span the exterior power. Hence $\Lambda^k(G_x)=\{0\}$. This holds for every point $x\in M$, so every fiber of $\Lambda^k G$ is the zero vector space. Therefore $\Lambda^k G$ is the zero vector bundle. [/guided] [/step] [step:Apply the rank bound to the bidegree factors] Assume first that $p>n$. Applying the previous step to $E=T^{*1,0}M$, which has rank $n$, gives \begin{align*} \Lambda^p E=0. \end{align*} Therefore \begin{align*} \Lambda^p E\otimes_{\mathbb C}\Lambda^q F=0, \end{align*} so its space of smooth sections is \begin{align*} \Gamma\left(M,\Lambda^p E\otimes_{\mathbb C}\Lambda^q F\right)=\{0\}. \end{align*} Thus $\Omega^{p,q}(M)=\{0\}$. If instead $q>n$, applying the same argument to $F=T^{*0,1}M$ gives $\Lambda^q F=0$, and hence again $\Lambda^p E\otimes_{\mathbb C}\Lambda^q F=0$. Therefore $\Omega^{p,q}(M)=\{0\}$. In either case, if $p>n$ or $q>n$, the space of forms of type $(p,q)$ is zero. This proves the theorem. [/step]