I tried this question twice on math.stackexchange but got no answer so I decided to move it here.

Let $ M$ be a smoth manifold. Then $ $ C^\infty(M):=\{f:M\longrightarrow \mathbb R; f\ \textrm{is smooth}\},$ $ is an $ \mathbb R$ -algebra with the pointwise product.

If you don’t know anything about smooth manifolds it doesn’t matter, all that you need to know is that $ C^\infty(M)$ is an $ \mathbb R$ -algebra for all that follows is purely algebraic.

Let us define the space of **vector fields** on $ M$ by: $ $ \mathfrak{X}(M):=\mathsf{der}_{\mathbb R}\ C^\infty(M):=\{X\in \mathsf{End}_{\mathbb R}(C^\infty(M)): X(fg)=fX(g)+X(f)g\}.$ $ This is a $ C^\infty(M)$ -module with $ $ (f\cdot X)(g):=f X(g),$ $

where $ fX(g)$ is the pointwise product of the functions $ f$ and $ X(g)$ .

We can then define the space of $ p$ -forms on $ M$ as the $ C^\infty(M)$ -module:

$ $ \Omega^p(M):=\mathsf{Hom}_{C^\infty(M)}(\Lambda^p \mathfrak{X}(M), C^\infty(M)).$ $ The **De Rham differential** on $ M$ is the degree one operator $ $ d: \Omega^p(M)\longrightarrow \Omega^{p+1}(M)$ $ given by $ $ \begin{eqnarray*} d\varepsilon(X_{1}, \ldots, X_{p+1})&&:=\sum_{\sigma\in\mathsf{Sh}(1, p)} \mathsf{sgn}(\sigma) X_{\sigma(1)}(\varepsilon(X_{\sigma(2)}, \ldots, X_{\sigma(p+1)})\ &&+\sum_{\sigma\in\mathsf{Sh}(2, p-1)}\mathsf{sgn}(\sigma) \varepsilon([X_{\sigma(1)}, X_{\sigma(2)}], X_{\sigma(3)}, \ldots, X_{\sigma(p+1)}), \end{eqnarray*}$ $ where $ [\cdot, \cdot]$ is the **commutator of vector fields** which is defined by $ $ [X, Y]:=X\circ Y-Y\circ X.$ $

**Notation:** For integers $ p, q\geq 1$ let us write $ S(p, q)$ as the subset of permutations $ \sigma$ of the set $ \{1, \ldots, p+q\}$ such that $ \sigma(1)<\ldots< \sigma(p)$ and $ \sigma(p+1)<\ldots< \sigma(p+q)$ . The elements of $ \mathsf{Sh}(p, q)$ are known as **$ (p, q)$ -shufles** for obvious resons.

Now, we can define a product $ $ \wedge: \Omega^p(M)\times \Omega^q(M)\longrightarrow \Omega^{p+q}(M)$ $ setting $ $ (\varepsilon\wedge \eta)(X_1, \ldots, X_{p+q}):=\sum_{\sigma\in\mathsf{Sh}(p, q)} \mathsf{sgn}(\sigma) \varepsilon(X_{\sigma(1)}, \ldots, X_{\sigma(p)}) \eta(X_{\sigma(p+1)}, \ldots, X_{\sigma(p+q)}).$ $ Can anyone help me to prove $ $ d(\varepsilon\wedge \eta)=d\varepsilon\wedge \eta+(-1)^p \varepsilon\wedge d\eta,$ $ for every $ \varepsilon\in \Omega^p(M)$ and $ \eta\in \Omega^q(M)$ ?

I know the property I want to show is a classical one but I can’t seem to find the proof using this algebraic formulation. I’ve already asked this question before and got no answer.

However, by that time things were more obscure so I decided to update with this improved version hoping someone could help me.

**Remark:** 1) The cardinality of the set $ \mathsf{Sh}(p, q)$ is $ \binom{p+q}{q}$ .

2) There is a bijection $ \mathsf{Sh}(p, q)\simeq \mathsf{Sh}(p-1, q)\sqcup \mathsf{Sh}(p, q-1)$ . This is clear since:

$ $ \mathsf{Sh}(p, q)=\{\sigma\in\mathsf{Sh}(p, q): \sigma(1)=1\}\sqcup\{\sigma\in\mathsf{Sh}(p, q): \sigma(p+1)=1\}.$ $ In particular:

$ $ \mathsf{Sh}(p+1, q)\simeq \mathsf{Sh}(p, q)\sqcup \mathsf{Sh}(p+1, q-1)\quad \textrm{and}\quad \mathsf{Sh}(p, q+1)\simeq \mathsf{Sh}(p, q)\sqcup \mathsf{Sh}(p-1, q+1).$ $

2) Heuristically, in light of $ 2)$ , when we add the terms $ d\varepsilon\wedge \eta$ and $ \varepsilon\wedge d\eta$ we get things which can be put into $ \sum_{\in \mathsf{Sh}(p, q)}$ which is desirable but the remaining terms $ \sum_{\sigma\in \mathsf{Sh}(p+1, q-1)}$ and $ \sum_{\sigma\in\mathsf{Sh}(p-1, q+1)}$ don’t cancel for the cardinality of the sets $ \mathsf{Sh}(p+1, q-1)$ and $ \mathsf{Sh}(p-1, q+1)$ are different. Indeed, there is no hope things will cancel when we do $ d\varepsilon\wedge \eta+\varepsilon\wedge d\eta$ for $ d\varepsilon\wedge \eta$ consists of terms which derivate $ \varepsilon(-)$ whereas $ \varepsilon\wedge d\eta$ consists of terms which derivate $ \eta(-)$ .