At 5:16 of this video, the following product rule is brought up:
$$ t_u( \alpha \wedge \beta)= (t_u \alpha) \wedge \beta + (-1)^{|\alpha|} \alpha \wedge t_u \beta$$
I want to know how to prove this.
I thought of proving this using the same logic for the exterior derivative:
$$ d( \alpha \wedge \beta) = d \alpha \wedge \beta + (-1)^{|\alpha|}\alpha \wedge d \beta$$
For a $p$ form $\alpha$ and a $q$ form $\beta$
Now the proof of this (at least one I know) proves it for $\alpha= \tilde{\alpha} dx_{i_1} \wedge dx_{i_2}...dx_{i_p}$ and $\beta=\tilde{\beta} dx_{j_1} \wedge dx_{j_2}... dx_{j_q}$ and the extends it to form fields with components along the different 'form basis' by the linearity of derivative.
I know the rule is linear, so I just need to figure out how to work it out with the a one component form.