[proofplan]
The proof is an operator-theoretic consequence of the definition of the $\bar{\partial}$-Neumann operator. The smooth bounded pseudoconvex hypotheses, together with closed range, are used to place us in the standard $L^2$ $\bar{\partial}$-Neumann setting where $N_q$ exists as the inverse of $\Box_q$ on the orthogonal complement of harmonic forms. Applying $\Box_qN_q=I-H_q$ and expanding the definition of $\Box_q$ gives the homotopy formula. For the solvability statement, the formula writes $f$ as an exact term plus a coexact term; the coexact term is orthogonal to $\ker \bar{\partial}_q$, while it also lies in $\ker \bar{\partial}_q$, so it must vanish.
[/proofplan]
[step:Expand $\Box_qN_q$ to obtain the homotopy identity]
Fix $f \in L^2_{(p,q)}(\Omega)$. By definition of the $\bar{\partial}$-Neumann operator,
\begin{align*}
N_q f \in \operatorname{Dom}(\Box_q)
\qquad\text{and}\qquad
\Box_q N_q f = f - H_q f.
\end{align*}
Since $N_q f \in \operatorname{Dom}(\Box_q)$, the two terms
\begin{align*}
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_q f
\qquad\text{and}\qquad
\bar{\partial}_q^*\bar{\partial}_qN_q f
\end{align*}
are well-defined elements of $L^2_{(p,q)}(\Omega)$. Expanding the definition of $\Box_q$ gives
\begin{align*}
f - H_q f
=
\Box_q N_q f
=
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_q f
+
\bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
Adding $H_q f$ to both sides yields
\begin{align*}
f
=
H_q f
+
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_q f
+
\bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
[guided]
We start from the defining property of the Neumann operator. The operator
\begin{align*}
N_q: L^2_{(p,q)}(\Omega) \to \operatorname{Dom}(\Box_q) \cap (\ker \Box_q)^\perp
\end{align*}
is chosen so that applying $\Box_q$ to $N_q f$ recovers the part of $f$ orthogonal to the harmonic forms:
\begin{align*}
\Box_q N_q f = f - H_q f.
\end{align*}
This already contains the desired identity; the only remaining step is to expand $\Box_q$.
Because $N_q f \in \operatorname{Dom}(\Box_q)$, the domain requirements in the definition of $\Box_q$ hold:
\begin{align*}
N_q f \in \operatorname{Dom}(\bar{\partial}_q) \cap \operatorname{Dom}(\bar{\partial}_{q-1}^*),
\end{align*}
with
\begin{align*}
\bar{\partial}_q N_q f \in \operatorname{Dom}(\bar{\partial}_q^*)
\qquad\text{and}\qquad
\bar{\partial}_{q-1}^*N_q f \in \operatorname{Dom}(\bar{\partial}_{q-1}).
\end{align*}
Therefore the two expressions
\begin{align*}
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_q f
\qquad\text{and}\qquad
\bar{\partial}_q^*\bar{\partial}_qN_q f
\end{align*}
are legitimate elements of $L^2_{(p,q)}(\Omega)$. Expanding
\begin{align*}
\Box_q
=
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*
+
\bar{\partial}_q^*\bar{\partial}_q
\end{align*}
at the vector $N_qf$ gives
\begin{align*}
f - H_q f
=
\Box_qN_q f
=
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_q f
+
\bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
After adding $H_qf$ to both sides, we obtain
\begin{align*}
f
=
H_q f
+
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_q f
+
\bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
[/guided]
[/step]
[step:Show that the coexact term vanishes for closed data orthogonal to harmonic forms]
Assume now that $f \in \operatorname{Dom}(\bar{\partial}_q)$, $\bar{\partial}_q f = 0$, and $H_q f = 0$. Define
\begin{align*}
u := \bar{\partial}_{q-1}^*N_q f \in L^2_{(p,q-1)}(\Omega).
\end{align*}
Since $N_q f \in \operatorname{Dom}(\Box_q)$, the definition of $\operatorname{Dom}(\Box_q)$ gives
\begin{align*}
u \in \operatorname{Dom}(\bar{\partial}_{q-1}).
\end{align*}
The homotopy identity reduces to
\begin{align*}
f
=
\bar{\partial}_{q-1}u
+
\bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
Define the coexact term
\begin{align*}
w := \bar{\partial}_q^*\bar{\partial}_qN_q f \in L^2_{(p,q)}(\Omega).
\end{align*}
Then
\begin{align*}
w = f - \bar{\partial}_{q-1}u.
\end{align*}
Since $\bar{\partial}_{q-1}u \in \operatorname{Range}(\bar{\partial}_{q-1})$, the distributional identity for the maximal closed $\bar{\partial}$-complex gives
\begin{align*}
\operatorname{Range}(\bar{\partial}_{q-1}) \subseteq \ker \bar{\partial}_q.
\end{align*}
Applying this inclusion to $u \in \operatorname{Dom}(\bar{\partial}_{q-1})$ yields
\begin{align*}
\bar{\partial}_q(\bar{\partial}_{q-1}u)=0.
\end{align*}
Together with $\bar{\partial}_q f=0$, this implies $w \in \ker \bar{\partial}_q$.
On the other hand, because $\bar{\partial}_qN_q f \in \operatorname{Dom}(\bar{\partial}_q^*)$, the element $w$ lies in $\operatorname{Range}(\bar{\partial}_q^*)$. Hence $w$ is orthogonal to $\ker \bar{\partial}_q$: for every $g \in \operatorname{Dom}(\bar{\partial}_q)$ with $\bar{\partial}_qg=0$,
\begin{align*}
(w,g)_{L^2}
=
(\bar{\partial}_q^*\bar{\partial}_qN_qf,g)_{L^2}
=
(\bar{\partial}_qN_qf,\bar{\partial}_qg)_{L^2}
=
0.
\end{align*}
Taking $g=w$ is allowed because $w \in \ker \bar{\partial}_q \subset \operatorname{Dom}(\bar{\partial}_q)$. Therefore
\begin{align*}
\|w\|_{L^2}^2 = (w,w)_{L^2}=0,
\end{align*}
so $w=0$. Consequently
\begin{align*}
f=\bar{\partial}_{q-1}u,
\end{align*}
with $u=\bar{\partial}_{q-1}^*N_qf$.
[guided]
Now assume that $f$ is $\bar{\partial}$-closed and has no harmonic part:
\begin{align*}
f \in \operatorname{Dom}(\bar{\partial}_q),
\qquad
\bar{\partial}_q f=0,
\qquad
H_qf=0.
\end{align*}
We define the candidate solution by
\begin{align*}
u := \bar{\partial}_{q-1}^*N_q f.
\end{align*}
This is the natural choice because the homotopy formula contains the exact term
\begin{align*}
\bar{\partial}_{q-1}\bar{\partial}_{q-1}^*N_qf.
\end{align*}
The domain condition is not automatic from notation alone, so we verify it: since $N_qf \in \operatorname{Dom}(\Box_q)$, the definition of $\operatorname{Dom}(\Box_q)$ gives
\begin{align*}
\bar{\partial}_{q-1}^*N_qf \in \operatorname{Dom}(\bar{\partial}_{q-1}).
\end{align*}
Thus $u \in \operatorname{Dom}(\bar{\partial}_{q-1})$, and $\bar{\partial}_{q-1}u$ is well-defined.
Because $H_qf=0$, the homotopy identity becomes
\begin{align*}
f
=
\bar{\partial}_{q-1}u
+
\bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
The only obstruction to concluding $\bar{\partial}_{q-1}u=f$ is the second term. Define
\begin{align*}
w := \bar{\partial}_q^*\bar{\partial}_qN_q f.
\end{align*}
Then
\begin{align*}
w = f - \bar{\partial}_{q-1}u.
\end{align*}
We prove $w=0$ by showing that $w$ is both in $\ker \bar{\partial}_q$ and orthogonal to $\ker \bar{\partial}_q$.
First, $w \in \ker \bar{\partial}_q$. Indeed, the hypothesis gives $\bar{\partial}_qf=0$. Also, because $u \in \operatorname{Dom}(\bar{\partial}_{q-1})$ and $\bar{\partial}_{q-1}u$ is an exact form, the distributional identity for the maximal closed $\bar{\partial}$-complex gives
\begin{align*}
\operatorname{Range}(\bar{\partial}_{q-1}) \subseteq \ker \bar{\partial}_q.
\end{align*}
Applying this inclusion to $u$ gives
\begin{align*}
\bar{\partial}_q(\bar{\partial}_{q-1}u)=0.
\end{align*}
Therefore
\begin{align*}
\bar{\partial}_q w
=
\bar{\partial}_q f
-
\bar{\partial}_q(\bar{\partial}_{q-1}u)
=
0.
\end{align*}
Second, $w$ is orthogonal to $\ker \bar{\partial}_q$. The reason is that $w$ lies in the range of the adjoint operator $\bar{\partial}_q^*$. More explicitly, since $\bar{\partial}_qN_q f \in \operatorname{Dom}(\bar{\partial}_q^*)$, for every $g \in \operatorname{Dom}(\bar{\partial}_q)$ with $\bar{\partial}_qg=0$, the Hilbert-space adjoint relation gives
\begin{align*}
(w,g)_{L^2}
=
(\bar{\partial}_q^*\bar{\partial}_qN_qf,g)_{L^2}
=
(\bar{\partial}_qN_qf,\bar{\partial}_qg)_{L^2}
=
0.
\end{align*}
Now apply this orthogonality to $g=w$. This is legitimate because the first part proved $w \in \ker \bar{\partial}_q$, hence $w \in \operatorname{Dom}(\bar{\partial}_q)$ and $\bar{\partial}_qw=0$. We obtain
\begin{align*}
\|w\|_{L^2}^2
=
(w,w)_{L^2}
=
0.
\end{align*}
Thus $w=0$ in $L^2_{(p,q)}(\Omega)$. The reduced identity is therefore
\begin{align*}
f=\bar{\partial}_{q-1}u,
\end{align*}
where
\begin{align*}
u=\bar{\partial}_{q-1}^*N_qf.
\end{align*}
This proves that $u$ solves the $\bar{\partial}$-equation with data $f$.
[/guided]
[/step]
