[proofplan]
We use the type decomposition of complex-valued differential forms and the decomposition $d=\partial+\bar\partial$ of the [exterior derivative](/theorems/1525). Since the exterior derivative satisfies $d^2=0$, expanding $(\partial+\bar\partial)^2$ gives three pieces of distinct bidegrees. The directness of the type decomposition then forces each bidegree component to vanish separately.
[/proofplan]
[step:Decompose the exterior derivative into its two type components]
Let $A^{p,q}(X)$ denote the smooth complex-valued differential forms of type $(p,q)$ on $X$. By [citetheorem:7004], the exterior derivative
\begin{align*}
d:A^{p,q}(X)\to A^{p+1,q}(X)\oplus A^{p,q+1}(X)
\end{align*}
has unique bidegree components
\begin{align*}
\partial:A^{p,q}(X)\to A^{p+1,q}(X)
\end{align*}
and
\begin{align*}
\bar\partial:A^{p,q}(X)\to A^{p,q+1}(X)
\end{align*}
such that $d=\partial+\bar\partial$ on complex-valued forms.
[guided]
Let $A^{p,q}(X)$ be the space of smooth complex-valued forms of type $(p,q)$ on the complex manifold $X$. The point of introducing this notation is that the exterior derivative does not preserve a single bidegree, but it changes bidegree in exactly two possible ways. By [citetheorem:7004], for every $\alpha\in A^{p,q}(X)$ one has
\begin{align*}
d\alpha\in A^{p+1,q}(X)\oplus A^{p,q+1}(X).
\end{align*}
Therefore there are unique operators
\begin{align*}
\partial:A^{p,q}(X)\to A^{p+1,q}(X)
\end{align*}
and
\begin{align*}
\bar\partial:A^{p,q}(X)\to A^{p,q+1}(X)
\end{align*}
such that
\begin{align*}
d\alpha=\partial\alpha+\bar\partial\alpha.
\end{align*}
This uniqueness is what will let us compare the different bidegree components after expanding $d^2\alpha$.
[/guided]
[/step]
[step:Expand the identity $d^2=0$ on a form of fixed type]
Fix integers $p,q\geq 0$ and let $\alpha\in A^{p,q}(X)$. The exterior derivative satisfies the standard identity $d^2=0$, so
\begin{align*}
0=d^2\alpha=d(d\alpha).
\end{align*}
Using $d=\partial+\bar\partial$ twice gives
\begin{align*}
0=(\partial+\bar\partial)(\partial\alpha+\bar\partial\alpha).
\end{align*}
By linearity of $\partial$ and $\bar\partial$,
\begin{align*}
0=\partial^2\alpha+\partial\bar\partial\alpha+\bar\partial\partial\alpha+\bar\partial^2\alpha.
\end{align*}
In this expression, $\partial\bar\partial\alpha$ means $\partial(\bar\partial\alpha)$ and $\bar\partial\partial\alpha$ means $\bar\partial(\partial\alpha)$.
[/step]
[step:Separate the expanded identity by bidegree]
The four terms in the expansion have the following types:
\begin{align*}
\partial^2\alpha\in A^{p+2,q}(X).
\end{align*}
\begin{align*}
\partial\bar\partial\alpha+\bar\partial\partial\alpha\in A^{p+1,q+1}(X).
\end{align*}
\begin{align*}
\bar\partial^2\alpha\in A^{p,q+2}(X).
\end{align*}
These three bidegrees are pairwise distinct. Since the decomposition of complex-valued forms into bidegrees is a direct sum, the equality
\begin{align*}
0=\partial^2\alpha+\partial\bar\partial\alpha+\bar\partial\partial\alpha+\bar\partial^2\alpha
\end{align*}
implies
\begin{align*}
\partial^2\alpha=0.
\end{align*}
\begin{align*}
\partial\bar\partial\alpha+\bar\partial\partial\alpha=0.
\end{align*}
\begin{align*}
\bar\partial^2\alpha=0.
\end{align*}
[guided]
We now use the direct-sum nature of the type decomposition. The operator $\partial$ raises the first bidegree by $1$ and leaves the second bidegree fixed, while $\bar\partial$ leaves the first bidegree fixed and raises the second bidegree by $1$. Therefore, starting from $\alpha\in A^{p,q}(X)$, the term $\partial^2\alpha$ has type $(p+2,q)$:
\begin{align*}
\partial^2\alpha\in A^{p+2,q}(X).
\end{align*}
The two mixed terms both have type $(p+1,q+1)$:
\begin{align*}
\partial\bar\partial\alpha\in A^{p+1,q+1}(X).
\end{align*}
\begin{align*}
\bar\partial\partial\alpha\in A^{p+1,q+1}(X).
\end{align*}
Finally, applying $\bar\partial$ twice gives type $(p,q+2)$:
\begin{align*}
\bar\partial^2\alpha\in A^{p,q+2}(X).
\end{align*}
Thus the expanded equation
\begin{align*}
0=\partial^2\alpha+\partial\bar\partial\alpha+\bar\partial\partial\alpha+\bar\partial^2\alpha
\end{align*}
is a sum of components lying in the three distinct subspaces $A^{p+2,q}(X)$, $A^{p+1,q+1}(X)$, and $A^{p,q+2}(X)$. Since the decomposition of forms by type is direct, a sum of components in distinct bidegrees can vanish only when each component vanishes. Hence
\begin{align*}
\partial^2\alpha=0.
\end{align*}
\begin{align*}
\partial\bar\partial\alpha+\bar\partial\partial\alpha=0.
\end{align*}
\begin{align*}
\bar\partial^2\alpha=0.
\end{align*}
This is the only place where the directness of the type decomposition is used: it converts one total-degree identity into three separate bidegree identities.
[/guided]
[/step]
[step:Conclude the operator identities on all bidegrees]
The integers $p,q\geq 0$ and the form $\alpha\in A^{p,q}(X)$ were arbitrary. Therefore, on every $A^{p,q}(X)$,
\begin{align*}
\partial^2=0.
\end{align*}
\begin{align*}
\bar\partial^2=0.
\end{align*}
\begin{align*}
\partial\bar\partial+\bar\partial\partial=0.
\end{align*}
These are precisely the asserted Dolbeault operator relations.
[/step]
