How to prove Leibniz rule for exterior derivative using abstract index notation I want to prove Leibniz rule for exterior derivative of wedge product using abstract index notation:

For $\omega\in \Omega^k(U),\eta\in\Omega^l(U)$, d$(\omega\wedge\eta)=\text{d}\omega\wedge\eta +(-1)^k\omega\wedge\text{d}\eta$.

My proof is given in the answer below.
 A: If you want a simple approach using the wedge product on general p-forms see below.  
The strategy is:


*

*the differential in $d$ gives you two expressions via the product rule.

*To operate with the wedge product on the second term, you need to reverse the wedge first, insert the $dx$ operator and re-reverse, this gives you the $-1$ term required.


Let $\alpha$ be an $n$-form, $\omega$ an $m$-form. $I,K$ sum over all basis combinations. 
$d(\alpha \land \omega)=d(\sum \alpha_I dx^I\land \sum \omega_K {dx}^K)\\ 
~~~~~~~~~~~~~~~=d(\sum \sum \alpha_I \omega_K dx^I \wedge dx^K) \\
~~~~~~~~~~~~~~~=\sum \sum ( \sum_i (\partial_i (\alpha_I) \omega_K (dx_i \land dx^I) \land dx^K+\alpha_I\ \partial_i \ \omega_K \ dx_i \land (-1)^{mn} dx^K \land dx^I    )) \\
~~~~~~~~~~~~~~~=\sum \sum ( \sum_i (\partial_i (\alpha_I) \omega_K (dx_i \wedge dx^I) \land dx^K\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+\alpha_I\ \partial_i\ \omega_K (-1)^{mn} (-1)^{mn}(-1)^n dx^I \land dx_i \land dx^K )) \\
~~~~~~~~~~~~~~~~=\sum \sum ( \sum_i (\partial_i\ (\alpha_I)\ \omega_K (dx_i \land dx^I) \land dx^K+(-1)^n\ \alpha_I\ \partial_i\ \omega_K\  dx^I \land dx_i \land dx^K\ )) \\
~~~~~~~~~~~~~~~~=\sum \sum ( d\ (\alpha_I)\ \omega_K\ dx^I \land dx^K+(-1)^n\ \alpha_I\ d \omega_K\  dx^I \land dx^K ) \\
~~~~~~~~~~~~~~~~= d\alpha \wedge \omega+(-1)^n \alpha \wedge d \omega $
