Question: I have to prove that $$\mathcal L_X\omega_{1}\wedge\omega_{2}=(\mathcal L_X\omega_{1})\wedge\omega_{2}+\omega_{1}\wedge(\mathcal L_X\omega_{2})$$ using the definition $$\mathcal L_X\omega=\frac{d}{dt}\bigg|_{t=0}\varphi_t^*\omega=\lim_{t\rightarrow 0} \frac{\varphi_t^*\omega-\omega}{t}$$
Attempt: I get $$\mathcal L_X(\omega_{1}\wedge\omega_{2})=\frac{d}{dt}\bigg |_{t=0}\varphi_t^*(\omega_{1}\wedge\omega_{2})$$
The homework hint says: Use $\varphi_t^*(\omega_{1}\wedge\omega_{2})=(\varphi_t^*\omega_{1})\wedge(\varphi_t^*\omega_{2})$ and the result is immediate, but I can't see the next step and I figure that thes proof hidden a trick.
Now use the product rule to calculate $\frac{d}{dt}\big\vert_{t=0}(\varphi_t^*\omega_1)\wedge(\varphi_t^*\omega_2)$.