How to see that $F$ $F$ dual is a surface term? The renormalisable 'theta term' that one can add to a Lagrangian describing Yang-Mills fields is often neglected on the grounds that it contributes a surface term. For QED, this is easy to see:
$$\theta \int F \wedge F = \theta \int \mathrm{d}(A \wedge F) $$
But for a non-Abelian field with $F = \mathrm{d}A + A \wedge A$, $F \wedge F$ contains an $A^4$ term which isn't obviously an exact form. Either I'm missing something obvious here, or perhaps $F \wedge F$ is not the correct way to write the theta term for a non-Abelian field?
EDIT (problem solved): I found a proof that 
$$ \mathrm{tr}(F \wedge F) = \mathrm{d}\, \mathrm{tr}\left(A \wedge \mathrm{d}A + \frac{2}{3}A^3 \right)$$
In section 10.5.5 (lemma 10.3) of Nakahara. As pointed out by ACuriousMind below, the proof isn't very enlightening, but one crucial step I was missing is that
$$ \mathrm{tr}(A^4) = 0$$
On account of the cyclicity of the trace and antisymmetry of the wedge product.
 A: So as to complement Nakahara's and Nogeira's proofs, when I did this calculation the most puzzling part was the origin of the $\frac{2}{3}$ in front of $A\wedge A\wedge A$, but it is easy to find it out:
\begin{align*}
   Tr[F\wedge F] =&\ Tr\left[dA\wedge dA+dA\wedge A\wedge A
                    +A\wedge A\wedge dA+A\wedge A\wedge A\wedge A\right] \\[0.5em]
                 =&\ Tr\left[dA\wedge dA+2dA\wedge A\wedge A\right]           \\
                 =&\ Tr\left[dA\wedge dA+\frac23(dA\wedge A\wedge A
                    -A\wedge dA\wedge A+A\wedge A\wedge dA)\right]      \\
                 =&\ Tr\left[d(A\wedge dA)+\frac23d(A\wedge A\wedge A)\right] \\
                 =&\ d\ Tr\left[A\wedge dA+\frac23A\wedge A\wedge A\right]
\end{align*}
where the key point is that $Tr[dA\wedge A\wedge A]=-Tr[A\wedge dA\wedge A]=Tr[A\wedge A\wedge dA]$, as can be easily checked from the cyclicity of the trace and  the antisymmetry of the wedge product (recall that $A$ are matrix-valued, not $\mathbb{R}$-valued 1-forms).
This shows that the Pontryagin density $Tr[F\wedge F]$ is exact and that its generating form is a Chern-Simons term.
A: In terms of the components $A=A_\mu dx^{\mu}$, we have$$
\\\
\frac{\theta}{2\pi}\mathrm{tr}\left[F\wedge F\right]=\frac{2\theta}{\pi}\mathrm{tr}\left[\varepsilon^{\mu\nu\rho\sigma}(\partial_{\mu} A_{\nu}+A_{\mu}A_{\nu})(\partial_{\rho} A_{\sigma}+A_{\rho}A_{\sigma})\right]
\\\\
$$
And then
$$
\frac{\theta}{2\pi}\mathrm{tr}\left[F\wedge F\right]=\frac{2\theta}{\pi}\mathrm{tr}\left[\varepsilon^{\mu\nu\rho\sigma}\partial_{\mu}(A_\nu\partial_\rho A_\sigma+\frac{2}{3}A_{\nu}A_\rho A_\sigma)\right]+\frac{2\theta}{\pi}\mathrm{tr}\left[A_{\mu}A_{\nu}A_\rho A_\sigma\right]\varepsilon^{\mu\nu\rho\sigma}
$$
by cyclic permutations on $\nu$, $\rho$, $\sigma$ and the fact that $\partial_\mu \partial_\nu$ is symmetric. Now, the last term vanish since a cyclic permutation of even number of elements is always odd (in this case four elements)
