Associativity of the exterior multiplication of forms. Let $\omega^{k}\in\bigwedge^{k}(V^{*})$ and $\eta^{l}\in\bigwedge^{l}(V^{*})$, be two exterior forms of degrees $k$ and $l$. The exterior product $\omega^{k}\wedge\eta^{l}$ is defined as $(k+l)$ form, by $$\omega^{k}\wedge\eta^{l}(v_{1},...,v_{k+l})=\dfrac{1}{k!l!}\sum_{\sigma\in S_{k+l}}{\operatorname{sgn}(\sigma)\omega^{k}(v_{\sigma(1)},...,v_{\sigma(k)})\eta^{l}(v_{k+1},...,v_{k+l})}$$
I wanna proof the exterior product has the following property: $(\omega^{k}\wedge\eta^{l})\wedge\mu^{m}=\omega^{k}\wedge(\eta^{l}\wedge\mu^{m})$
My approach: Note that, $\omega^{k}\wedge\eta^{l}$ is a multilinear and antisymmetric map acting on $(k+l)$ vectors. Then $$((\omega^{k}\wedge\eta^{l})\wedge\mu^{m})(v_{1},..v_{k+l+m})=$$
$$\dfrac{1}{(k+l)!m!}\sum_{\sigma\in S_{k+l+m}}{\operatorname{sgn}(\sigma)(\omega^{k}\wedge\eta^{l})(v_{\sigma(1)}.,v_{\sigma(k+l)})\mu^{m}(v_{k+l+1},.,v_{k+l+m})}$$
And now, descompose the permutation group $S_{k+l+m}$ into residual classes with respect to the subgroup $S_{k+l}\subset S_{k+l+m}$ formed by all permutations acting as the identity on the last m indices, then each residue class R thus consists of all permutations $\sigma\in S_{k+l+m}$ with fixed values $\sigma(k+l+1),...,\sigma(k+l+m)$, but how I can continuous?
 A: Let us call $\mathcal C$ the set of these residue classes. Note that each $R\in\mathcal C$ is isomorph to $S_{k+l}$. Choose one class $R$ and, within $R$, a particular permutation $\sigma_R$. Each element $\sigma\in R$ can be decomposed as $\sigma=\sigma_R\circ \pi$ where $\pi\in S_{k+l}$. Then:
$$
\DeclareMathOperator{\sgn}{sgn}
\begin{multline}
((w\wedge\eta)\wedge \mu)(v_1,\dots v_{k+l+m}) =\frac{1}{(k+l)!m!}\sum_{R\in\mathcal C} \sgn(\sigma_R)\left(\sum_{\sigma\in R}\sgn(\pi)\,(w\wedge\eta)(v_{\sigma(1)},\dots v_{\sigma(k+l)})\right) \\
 \mu(v_{\sigma_R(k+l+1)},\dots v_{\sigma_R(k+l+m)})
\end{multline}
$$
Now, note that all terms in the large parenthesis are equal, because they are permutations $\pi$ from a fixed ordering given by $\sigma_R$. Since $R$ is isomorph to $S_{k+l}$, there are $(k+l)!$ such terms, so:
$$
\DeclareMathOperator{\sgn}{sgn}
\begin{multline}
((w  \wedge\eta)\wedge \mu)  (v_1,\dots v_{k+l+m}) =\frac{1}{m!}\sum_{R\in\mathcal C}\sgn(\sigma_R) \,(w\wedge\eta)(v_{\sigma_R(1)},\dots v_{\sigma_R(k+l)})\,\mu(v_{\sigma_R(k+l+1)},\dots v_{\sigma_R(k+l+m)}) \\
=\frac{1}{k!l!m!}\sum_{R\in\mathcal C}\sgn(\sigma_R)\sum_{\tau\in S_{k+l}}  \sgn(\tau)\,w(v_{\tau(\sigma_R(1))},\dots v_{\tau(\sigma_R(k))})  \, \eta(v_{\tau(\sigma_R(k+1))},\dots v_{\tau(\sigma_R(k+l))}) \\\mu(v_{\sigma_R(k+l+1)},\dots v_{\sigma_R(k+l+m)})
\end{multline}
$$
But, again, all permutations $\sigma\in S_{k+l+m}$ can be decomposed as $\sigma=\tau\circ\sigma_R$ and, since $\tau$ acts as the identity on the last $m$ indices, we have $\sigma=\sigma_R$ for these indices used in the arguments of $\mu$. Thus we finally obtain:
$$
\DeclareMathOperator{\sgn}{sgn}
\begin{multline}
((w\wedge\eta)\wedge \mu)(v_1,\dots v_{k+l+m}) =\frac{1}{k!l!m!}\sum_{\sigma\in S_{k+l+m}}\sgn(\sigma)\,w(v_{\sigma(1)},\dots v_{\sigma(k)})  \, \eta(v_{\sigma(k+1)},\dots v_{\sigma(k+l)}) \\\mu(v_{\sigma(k+l+1)},\dots v_{\sigma(k+l+m)})
\end{multline}
$$
and the product does not depend on the order we associate the operations. $\square$
NOTE: By your wording I guess you are (or were, sorry I am late) reading Global Analysis by Agricola and Friedrichs. I am with it now :)
