Lie derivative commutes with interior product

Let $i_X : \Omega^k M \to \Omega^{k-1} M$ be the interior product for a smooth manifold $M$ and a smooth vector field $X$ with flow $\Phi$ and $$L_X : \Omega^k \to \Omega^k, \; \omega \mapsto L_X \omega := \frac{\text d}{\text d t} \Big| _{t=0} \Phi^*_t \omega$$ the Lie derivative of a $k$-form. Show that $$i_X \circ L_X = L_X \circ i_X.$$

To be honest, I have no clue. What I have proven so far is that the Lie derivative commutes with the exterior derivative ($d \circ L_X = L_X \circ d$), and I have also proven Cartan's magic formula ($L_X = i_X \circ d + d \circ i_X$) by induction. Is there an easy way to use both of these to prove the upper statement?

$i_X\circ L_X=i_X\circ (i_X\circ d+d\circ i_X)=i_X\circ d\circ i_X$ since $i_X\circ i_X=0$.
$L_X\circ i_X=(i_X\circ d +d\circ i_X)\circ i_X=i_X\circ d\circ i_X$. You deduce that $i_X\circ L_X=L_X\circ i_X$.
• Oh, I didn't know that $i_X^2 = 0$... But that is obvious. Thank you. – Yusel Jul 10 '17 at 17:25