If $A$ is the alternating operator, prove that $A(A(\omega\otimes\eta)\otimes\nu)=A(\omega\otimes\eta\otimes\nu)$ Let $V$ be a vector space over $\mathbb{R}$. A $p$-tensor on $V$ is a multilinear map 
    $$
 \omega: \underbrace{V\times\cdots\times V}_\text{$p$ times}\longrightarrow\mathbb{R}
$$
The set of all $p$-tensors is denoted $\mathcal{T}^p(V)$.
Let $\omega\in\mathcal{T}^p(V)$, and define $A:\mathcal{T}^p(V)\longrightarrow \Lambda^p(V)$ by
    $$
 A(\omega)(v_1,\ldots,v_p) = \frac{1}{p!}\sum_{\sigma\in S_p}{(-1)^\sigma\omega(v_{\sigma(1)},\ldots,v_{\sigma(p)})}
$$
Finally, let $\omega\in\mathcal{T}^p(V)$ and $\eta\in\mathcal{T}^q(V)$, and define the tensor product $\omega\otimes\eta\in\mathcal{T}^{p+q}(V)$
    $$
 \omega\otimes\eta(v_1,\ldots,v_p,v_{p+1},\ldots,v_{p+q}) = \omega(v_1,\ldots,v_p)\eta(v_{p+1},\ldots,v_{p+q})
$$
I'm trying to prove the following identity:
$$
A(A(\omega\otimes\eta)\otimes\nu) = A(\omega\otimes A(\eta\otimes\nu)) = A(\omega\otimes\eta\otimes\nu)
$$
Here was my initial approach:
Let $l=p+q+r$ and $m=p+q$.
        \begin{align*} 
     A(A(\omega\otimes\eta)&\otimes\nu)(v_1,\ldots,v_l) = \frac{1}{l!}\sum_{\rho\in S_{l}}{(-1)^\rho A(\omega\otimes\eta)(v_{\rho(1)},\ldots,v_{\rho(m)})\nu(v_{\rho(m+1)},\ldots, v_{\rho(l)})}\\
         &= \frac{1}{l!m!}\sum_{\rho\in S_{l}}{\sum_{\sigma\in S_m}(-1)^{\sigma\rho}\omega(v_{\sigma\rho(1)},\ldots,v_{\sigma\rho(p)})\eta(v_{\sigma\rho(p+1)},\ldots,v_{\sigma\rho(m)})\nu(v_{\rho(m+1)},\ldots, v_{\rho(l)})}\, .
    \end{align*}
Let $H\subset S_l$ be a set of all permutations such that for any $g,h\in H$, $g(i) = h(i)$ is not true for all $i=m,\ldots,l$. Then, any element of $S_l$ can be written as $\rho h$ for some $\rho\in S_m$. This is equivalent to representing a permutation of $S_l$ as first selecting $r$ ordered elements from $\{1,\ldots,l\}$, placing them at the end of a list and then permuting the remaining $m$ elements and placing them at the start. The total number of ways to do this is equal to $(l!(l-r)!)m!=l!=|S_l|$. Thus the above sum can be rewritten as 
\begin{align*}
     \frac{1}{l!m!}\sum_{h\in H}(-1)^h\nu(v_{h(m+1)},\ldots,v_{h(l)})\Bigg({\sum_{\rho,\sigma\in S_m}{(-1)^{\sigma\rho h}\omega(v_{\sigma\rho h(1)},\ldots,v_{\sigma\rho h(p)})\omega(v_{\sigma\rho h(p+1)},\ldots,v_{\sigma\rho h(m)})}\Bigg)}
    \end{align*}
I should be able to reduce the second sum to just 
$$
m!\sum_{\rho\in S_m}{(-1)^{\rho}\omega(v_{\sigma\rho h(1)},\ldots,v_{\sigma\rho h(p)})\omega(v_{\sigma\rho h(p+1)},\ldots,v_{\sigma\rho h(m)})}
$$
Which would give me the desired result but I can't seem to figure out how. Any help would be appreciated.
 A: Firstly, you have to be a little more careful with the indices. 
Let $w_i = v_{\rho(i)}$ for $1\le i\le m$. Then
$$
\begin{align*}
\newcommand{\tens}{\otimes}
A(\omega\tens\eta)(w_1,\ldots,w_m)
=&
\frac{1}{m!}
\sum_{\sigma\in S_m}
(-1)^\sigma
\omega(w_{\sigma(1)},\ldots,w_{\sigma(p)})
\eta(w_{\sigma(p+1)},\ldots,w_{\sigma(m)})
\\
=&
\frac{1}{m!}
\sum_{\sigma\in S_m}
(-1)^\sigma
\omega(v_{\rho\sigma(1)},\ldots,v_{\rho\sigma(p)})
\eta(v_{\rho\sigma(p+1)},\ldots,v_{\rho\sigma(m)}).
\end{align*}
$$
Note that this results in a different order of composition of $\rho$ and $\sigma$ in the index. This is crucial.
Now we can do the following to simplify the double sum.
The key steps, are noticing that for $i>m$, $\sigma(i)=i$, so $\rho\sigma(i)=\rho(i)$, giving the first equality below, and that we can then reindex by summing over $\tau = \rho\sigma$. 
$$
\begin{align*} 
\frac{1}{l!m!}
\sum_{\rho\in S_{l}}&
\sum_{\sigma\in S_m}(-1)^{\sigma\rho}
\omega(v_{\rho\sigma(1)},\ldots,v_{\rho\sigma(p)})
\eta(v_{\rho\sigma(p+1)},\ldots,v_{\rho\sigma(m)})
\nu(v_{\rho(m+1)},\ldots, v_{\rho(l)})
\\
=&
\frac{1}{l!m!}
\sum_{\rho\in S_l}
\sum_{\sigma\in S_m}
(-1)^{\sigma\rho}
\omega(v_{\rho\sigma(1)},\ldots,v_{\rho\sigma(p)})
\eta(v_{\rho\sigma(p+1)},\ldots,v_{\rho\sigma(m)})
\nu(v_{\rho\sigma(m+1)},\ldots, v_{\rho\sigma(l)})
\\
=&\newcommand{\inv}{^{-1}}
\frac{1}{l!m!}
\sum_{\sigma\in S_m}
\sum_{\substack{\tau\in S_l\\\rho:=\tau\sigma^{-1}}}
(-1)^{\sigma\tau\sigma\inv}
\omega(v_{\tau\sigma\inv\sigma(1)},\ldots,v_{\tau\sigma\inv\sigma(p)})
\eta(v_{\tau\sigma\inv\sigma(p+1)},\ldots,v_{\tau\sigma\inv\sigma(m)})
\nu(v_{\tau\sigma\inv\sigma(m+1)},\ldots, v_{\tau\sigma\inv\sigma(l)})
\\
=&
\frac{1}{l!m!}
\sum_{\sigma\in S_m}
\sum_{\tau\in S_l}
(-1)^{\tau}
\omega(v_{\tau(1)},\ldots,v_{\tau(p)})
\eta(v_{\tau(p+1)},\ldots,v_{\tau(m)})
\nu(v_{\tau(m+1)},\ldots, v_{\tau(l)})
\\
=&
\frac{m!}{l!m!}
\sum_{\tau\in S_l}
(-1)^{\tau}
\omega(v_{\tau(1)},\ldots,v_{\tau(p)})
\eta(v_{\tau(p+1)},\ldots,v_{\tau(m)})
\nu(v_{\tau(m+1)},\ldots, v_{\tau(l)})
\\
=&
A(\omega\tens\eta\tens\nu)(v_1,\ldots,v_l)
\end{align*}
$$
