Exact real $(1,1)$-form on a Kähler manifold is $d^c$-closed. I am trying to prove a version of the global $dd^c$-lemma as a corollary of the following proposition:

Let $M$ a closed and Kähler manifold. Let $\alpha \in \Omega^q(M)$ a $d$-closed form and a $d^c$-exact form. Then there exists $\beta \in \Omega^{q-2}(M)$ such that $\alpha = dd^c\beta$.

I am trying to apply this result to:

Let $M$ a closed Kähler manifold and $\alpha \in \Omega^{1,1}(M)$ real and $d$-exact form. Then there exists $\phi \in C^{\infty}(M)$ such that $\alpha = dd^c\phi$.

Is a immediate consequence of the proof of the first claim that if I show that $\alpha$ is $d$-exact and $d^c$-closed then the theorem holds and I get the result. I am having trouble to show that under the hypothesis $\alpha \in \Omega^{1,1}(M)$ real and $d$-exact then $\alpha$ is $d^c$-closed.
Could someone help me?
Thanks.
 A: Recall that $d = \partial + \bar{\partial}$ and $d^c = -i(\partial - \bar{\partial})$.
Let $\alpha$ be a $d$-exact, and hence $d$-closed, real $(1, 1)$-form. Then $\alpha = d\eta$ for some one-form $\eta$. If $\eta^{1,0}$ and $\eta^{0,1}$ denote the $(1, 0)$ and $(0, 1)$ components of $\eta$ respectively, then we have $\eta = \eta^{1,0} + \eta^{0,1}$ and 
$$\alpha = d\eta = (\partial + \bar{\partial})(\eta^{1,0} + \eta^{0,1}) = \partial\eta^{1,0} + (\bar{\partial}\eta^{1,0} + \partial\eta^{0,1}) + \bar{\partial}\eta^{0,1}.$$
Comparing bidegrees, we see that
\begin{align*}
\partial\eta^{1,0} &= 0\\
\bar{\partial}\eta^{1,0} + \partial\eta^{0,1} &= \alpha\\
\bar{\partial}\eta^{0,1} &= 0.
\end{align*}
Now consider the one-form $\eta' := -i(\eta^{1,0} - \eta^{0,1})$. Note that
\begin{align*}
d^c\eta' &= -id^c(\eta^{1,0} - \eta^{0,1})\\
&= -(\partial - \bar{\partial})(\eta^{1,0} - \eta^{0,1})\\ 
&= -\partial\eta^{1,0} + (\bar{\partial}\eta^{1,0} + \partial\eta^{0,1}) - \bar{\partial}\eta^{0,1}\\  
&= \alpha,
\end{align*}
so $\alpha$ is also $d^c$-exact.
As $\alpha \in \Omega^2(M)$ is $d$-closed and $d^c$-exact, there is $\beta \in \Omega^0(M)$ such that $\alpha = dd^c\beta$ - here $\beta$ is nothing but a smooth complex-valued function. As $d$ and $d^c$ are real operators and $\alpha$ is real, 
$$\alpha = \overline{\alpha} = \overline{dd^c\beta} = dd^c\bar{\beta}$$
so letting $\phi = \operatorname{Re}\beta = \frac{1}{2}(\beta + \bar{\beta})$, we see that $\phi$ is a smooth real-valued function on $M$ (i.e. $\phi \in C^{\infty}(M)$) such that $\alpha = dd^c\phi$.
