The fact that the Lie derivative $\mathcal{L}_X$ commutes with the exterior derivative $d$ says that $\mathcal{L}_X$ induces a chain map on $\Omega^*(M)$, the de Rham complex.
The Cartan magic formula says that $\mathcal{L}_X$ is trivial on cohomology, being chain homotopic to the $0$ map (with chain homotopy given by contraction with $X$).
I have little intuition to the Cartan magic formula: its proof (for me), while simple after the necessary preliminaries, does not show the formula to be self-evident or self-suggestive.
However, if there were some reason to expect that $\mathcal{L}_X$ is trivial on cohomology beforehand, we could hope that it may come from a chain homotopy, for which one very natural map which presents itself as a candidate for the homotopy is the contraction, since it is a natural map which lowers degree. Therefore, the formula would suggest itself for investigation and now it would be a matter of checking our hopes.
My question, therefore, is: Is there some reason why we would expect $\mathcal{L}_X$ to induce a trivial map in cohomology from its definition?
The exponential of the Lie derivative describes flow along $X$, which is a one-parameter family of diffeomorphisms. So all of them are homotopic to the identity, and hence all induce the identity on cohomology. Moreover the homotopy is given by flowing backwards.