How can I prove that interior product obeys a graded Leibniz rule? I want to prove that $i_{X}(\omega\wedge\phi)=i_{X}\omega\wedge\phi+(-1)^{k}\omega\wedge i_{X}\phi.$
I was thinking I many be able to adapt the proof that the exterior derivative obeys the graded Leibniz rule. Failing that I have no idea how to prove this.
 A: You can also do it directly from the definitions. First recall that if $\alpha, \beta$ are alternating $k$ and $l$ forms, then $$\alpha\wedge\beta(v_1,\dots,v_{k+l})=\sum_{\sigma \in(k,l)}\text{sgn}(\sigma)\alpha(v_{\sigma(1)},\dots v_{\sigma(k)})\beta(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})$$ where by the notation $\sigma \in(k,l)$ I mean that $\sigma\in S^{k+l}$ is such that $\sigma(1)<\dots<\sigma(k)$ and $\sigma(k+1)<\dots<\sigma(k+l)$. This is nothing substantial, it is just useful to reduce the mess by a tiny bit.
Now we are ready to expand the definitions. Call $X=v_1$, this will be helpful.
$$i_{v_1}(\omega\wedge\phi)(v_2,\dots,v_{k+l})=(\omega\wedge\phi)(v_1,\dots,v_{k+l})=\sum_{\sigma \in(k,l)}\text{sgn}(\sigma)\omega(v_{\sigma(1)},\dots v_{\sigma(k)})\phi(v_{\sigma(k+1)},\dots,v_{\sigma(k+l)})=\sum_{\sigma(1)=1}+\sum_{\sigma(k+1)=1}=\sum_{\tau\in (k-1,l)}\text{sgn}(\tau)\omega(v_{\tau(1)},\dots v_{\sigma(k)})\phi(v_{\tau(k+1)},\dots,v_{\tau(k+l)})+\sum_{\tau\in (k,l-1)}\text{sgn}(\tau)\omega(v_{\tau(1)},\dots v_{\tau(k)})\phi(v_{\tau(k+1)},\dots,v_{\tau(k+l)}).$$
Now in the first sum $\tau$ permutes $2,\dots, k+l$ and has the same sign as the corresponding $\sigma$, whereas in the second case it permutes $1,\dots k+l-1$ and has sign $(-1)^k$ (to put $k+1$ in first place it takes $k+1-1=k$ moves). Now the last sum is exactly $$i_{v_1}\omega\wedge\phi(v_2,\dots,v_{k+l})+\omega\wedge i_{v_1}\phi(v_2,\dots, v_{k+l}).$$
