Maxwell equations of motion from $S = \frac{-1}{2} \int F \wedge \ast F$ 
I'm trying to understand the following equation, used in the derivation of the equations of motion. Let $S = \frac{-1}{2} \int F \wedge \ast F$ and $F = dA$. Let $\delta$ denote variation. Then
$$\delta S = - \int d \delta A \wedge \ast F.$$
(In the full derivation we then int. by parts and get $d\ast F = 0$ for the EOM with no currents.)

In tensor notation, $F = F_{\mu\nu}$ and (I think) $(\ast F) = (\ast F)_{\gamma \delta} = \sqrt{|g|} \epsilon^{\alpha\beta}_{\,\,\,\,\,\, \gamma \delta} F_{\alpha \beta}$ (where $|g|$ is the determinant of the metric). The wedge product is antisymmetric, so (I think) $F\wedge \ast F = \epsilon^{\mu\nu \gamma\delta} F_{\mu\nu} (\ast F)_{\gamma \delta}$. After these considerations, I do not see how we end up with the above result; I'd expect two terms from the product rule for variations, for instance, and where did the factor of $\frac12$ go?
 A: Actually, once being in differential form formalism, the best is to stay in it and thereby benefit from it. 
In this sense let's evaluate the variation of the EM-action using F=dA:
$$\delta S =-\frac{1}{2}\delta\int_V dA \wedge\star dA = -\frac{1}{2}\int_V(d\delta A\wedge \star dA + dA \wedge \star d \delta A)=-\frac{1}{2}\int_V 2 d\delta A\wedge \star dA$$
Here we used a hodge operator rule for forms $\lambda, \omega  \in \Lambda^k$
 for $\lambda = d\delta A$ and $\omega=dA$:
$$\lambda \wedge \star \omega = (-1)^q <\lambda,\omega>e = (-1)^q <\omega,\lambda>e =\omega\wedge \star \lambda$$
with $$e: = e_1 \wedge e_2 \wedge \ldots \wedge e_n \in \Lambda^n$$ where $e_i$ form an orthogonal base in $\Lambda$ and the index $q$ is related with the signature of the metric of the used space. The braket pair $<a,b>$ represents the scalar product on the space $\Lambda^k$.
Second step (product rule): 
$$ d(\delta A\wedge\star dA) = d\delta A\wedge\star dA -\delta A\wedge d\star dA$$
It follows: 
$$ -d\delta A\wedge\star d A = -d(\delta A\wedge\star dA)-\delta A\wedge d\star dA$$
Therefore we get for the varied action integral: 
$$\delta S = \int_V d(\delta A\wedge\star dA) + \int_V \delta A\wedge d\star dA $$
We transform the first volume integral into an integral over the volume's surface according to Stokes theorem and realize that the variations $\delta A$ vanish on the surface:
$$0=\delta S = \int_{\partial V}\delta A\wedge\star dA + \int_V \delta A\wedge d\star dA = \int_V \delta A\wedge d\star dA$$
As the integral $$\int_V \delta A\wedge d\star dA=0$$ for all variations $\delta A$ we can conclude that: 
$$ d\star dA=0 $$ which corresponds to the second part of the free Maxwell equations.
