Is the Lie derivative $L_{X}(\omega \wedge \mu)$ an exact form? 
Let $\omega$ be an $n$-form and $\mu$ be an $m$-form where both are acting on a manifold $M$. Is the Lie derivative $L_{X}(\omega \wedge \mu)$ where $X$ is a smooth vector field acting on $M$ an exact form? 

I think it is but I've been unable to prove it, so any help would be greatly appreciated. 
 A: Hint Use Cartan's Magic Formula, which says that the Lie derivative $\mathcal L_X$ of a differential form $\alpha$ satisfies
$$\mathcal L_X \alpha = \iota_X d \alpha + d (\iota_X \alpha) .$$
From the statement of the original problem in the comments, $L_X(\omega \wedge \mu)$ can be integrated, so $L_X(\omega \wedge \mu)$, and hence $\alpha := \omega \wedge \mu$, is a top form. In particular, $d \alpha = 0$; so, $\mathcal L_X \alpha$ is exact (and we can finish proving the claim in the comments with Stokes' Theorem).
NB that since we used the fact that $\omega \wedge \mu$ is a top form, this argument doesn't apply to the question in the full stated generality (i.e., to forms of general degree), and indeed, one can readily construct a counterexample using the above identity as a guide.
A: In general this is not true. Recall that
$$
 L_X(\omega)  = i_x d\omega + d i_x\omega
$$
where you see that the right part is exact and the left part mustn't be. As an example for your case take $N$ a manifold with a non exact form $\mu$
and let $\omega$ be a 0-form (function) on $\mathbb{R}$ and define $M=N\times\mathbb{R}$ and note that the extension of $\mu$ is still not exact. Taking $\omega=x$ and $X=\partial_x$ where $x$ is the variable in $\mathbb{R}$ one sees
$$
 L_X(\omega\wedge \mu) = L_X(\omega)\wedge\mu +(-1)^0\omega\wedge L_X(\mu)=\mu
$$
since the projection of $X$ on $T_N$ vanishes.
