# Leibniz rule for exterior derivative of wedge product

I'm trying to show $$\text{d}(\alpha\wedge\gamma)=\text{d}\alpha\wedge\gamma+(-1)^p\alpha\wedge \text{d}\gamma$$ for all $$p$$-forms $$\alpha$$ and $$\gamma$$, $$\text{d}$$ is exterior derivative. I want to know the proof using the following definition in book The Road to Reality by Penrose:

The coordinate definition of the exterior derivative of the $$p$$-form $$\alpha$$ is a $$(p+1)$$-form that is written $$\text{d}\alpha$$, which has components \begin{align*} (\text{d}\alpha)_{ab\ldots i}=\frac{\partial}{\partial x^{[a}}\alpha_{bc\ldots i]}, \end{align*} (The notation looks a bit awkward here. The antisymmetrization---which is the key feature of the expression---extends across all $$p+1$$ indices, including the one on the derivative symbol.)

Here's my attempt at a proof, which is partially based off of a proof that I found online. Using the above definition for the component of a differential, we have \begin{align*} \text{d}(\alpha\wedge\gamma)_{ab\ldots i}&=\frac{\partial}{\partial x^{[a}}(\alpha\wedge\gamma)_{bc\ldots i]}\\ &=\frac{\partial}{\partial x^{[a}}\alpha_{[bc\ldots e}\gamma_{fg\ldots i]]} \end{align*} It's worth mentioning that the wedge product can very neatly be written using antisymmetrization of the components of $$\alpha$$ and $$\gamma$$ like so: \begin{align*} (\alpha\wedge\gamma)_{bc\ldots i}=\alpha_{[bc\ldots e}\gamma_{fg\ldots i]} \end{align*} This next step was the part that I borrowed off the internet: the inner brackets is redundant here because composition of permutations results in another permutation of indices. \begin{align*} \frac{\partial}{\partial x^{[a}}\alpha_{[bc\ldots e}\gamma_{fg\ldots i]]}=\frac{\partial}{\partial x^{[a}}\alpha_{bc\ldots e}\gamma_{fg\ldots i]} \end{align*} Using the chain rule, \begin{align*} \frac{\partial}{\partial x^{[a}}\alpha_{bc\ldots e}\gamma_{fg\ldots i]}=\frac{\partial\alpha_{bc\ldots e}}{\partial x^{[a}}\gamma_{fg\ldots i]}+\alpha_{bc\ldots e}\frac{\partial\gamma_{fg\ldots i]}}{\partial x^{[a}} \end{align*} I'm uncertain about the validity of the next part: the index $$a$$ been shifted to be between the indices $$bc\ldots e$$ and $$fg\ldots i$$. We have \begin{align*} \text{d}(\alpha\wedge\gamma)_{ab\ldots i}&=\frac{\partial\alpha_{bc\ldots e}}{\partial x^{[a}}\gamma_{fg\ldots i]}+\alpha_{[bc\ldots e}\frac{\partial\gamma_{fg\ldots i]}}{\partial x^{a}}\\ &=\frac{\partial\alpha_{bc\ldots e]}}{\partial x^{[[a}}\gamma_{fg\ldots i]}+\alpha_{[bc\ldots e}\frac{\partial\gamma_{fg\ldots i]]}}{\partial x^{[a}}\\ &=(\text{d}\alpha)_{[abc\ldots e}\gamma_{fg\ldots i]}+\alpha_{[bc\ldots e}(\text{d}\gamma)_{afg\ldots i]}\\ &=(\text{d}\alpha\wedge\gamma)_{ab\ldots i}+(\alpha\wedge \text{d}\gamma)_{bc\ldots eaf\ldots i} \end{align*} The last step is shifting the index $$a$$ over to the start of the subscript which results in a coefficient of $$(-1)^p$$ because of that is the order of permutation if there are $$p$$ element to the left of $$a$$. \begin{align*} (\text{d}\alpha\wedge\gamma)_{ab\ldots i}+(\alpha\wedge \text{d}\gamma)_{bc\ldots eaf\ldots i}&=(\text{d}\alpha\wedge\gamma)_{ab\ldots i}+(-1)^p(\alpha\wedge d\gamma)_{ab\ldots i}\\ &=(\text{d}\alpha\wedge\gamma+(-1)^p \alpha\wedge \text{d}\gamma)_{ab\ldots i} \end{align*} So is my proof valid, or incorrect even though the conclusion is correct?

Thank you.