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A pure-type $d$-closed form is both $\partial$-closed and $\bar{\partial }$-closed, because these two derivatives have different types. Let $k=p+q$. The smooth Poincare lemma on the polydisc gives $\eta \in {\mathcal{A}}_{\mathbb{C}}^{k-1}$ with $d\eta =\alpha$. We will modify $\eta$ by exact terms without changing $d\eta$.

Write $\eta ={\sum }_r{\eta }^{r,k-1-r}$. Starting with the smallest holomorphic degree, eliminate every component with $r<p-1$. Once the lower components are zero, the $(r,k-r)$ component of $d\eta$ says $\bar{\partial }{\eta }^{r,k-1-r}=0$, since $r<p$. Its antiholomorphic degree is positive, so the Dolbeault-Poincaré lemma supplies ${\nu }^{r,k-2-r}$ with $\bar{\partial }\nu ={\eta }^{r,k-1-r}$. Replace $\eta$ by $\eta -d\nu$: this removes the component at $r$ and changes only the next holomorphic degree.

Next, work downwards from the largest holomorphic degree and eliminate the components with $r>p$. Once the higher components are zero, the corresponding component of $d\eta$ gives $\partial {\eta }^{r,k-1-r}=0$. By part (b), write this as $\partial {\nu }^{r-1,k-1-r}$ and again subtract $d\nu$. This affects only the next lower holomorphic degree. The two finite procedures leaveThe components of $d\eta =\alpha$ outside type $(p,q)$ now give $\bar{\partial }u=0$ and $\partial v=0$, while its $(p,q)$ component gives $\alpha =\partial u+\bar{\partial }v$.

Since $q\ge 1$, the Dolbeault-Poincaré lemma gives $u=\bar{\partial }a$ with $a$ of type $(p-1,q-1)$. Since $p\ge 1$, part (b) gives $v=\partial b$ of the same type. The anticommutation identity therefore yieldsThis proves the Bott-Chern Poincaré lemma using precisely the permitted primitives: