If $v_i=\sum a_{ij}e_j$, then $\nu (v_1, ..., v_n)= \det(a_{ij})=\mathscr{vol}(v_1, ... v_n).$ Let $\nu$ be the $n-$form in $\mathbb{R}^n$ defined by
$$\nu(e_1, e_2, ..., e_n)=1,$$
where $\{e_i\}$,  $i=1, ..., n$, is the canonical basis of $\mathbb{R}^n$. Show that:

*

*If $v_i=\sum a_{ij}e_j$, then

$$\nu (v_1, ..., v_n)= \det(a_{ij})=\mathscr{vol}(v_1, ... v_n).$$
(the form $\nu$ is called the volume element of $\mathbb{R}^n$)


*$\nu=dx_1\wedge dx_2\wedge...\wedge dx_n$.

I am thinking of resolving this question by using item 2 before item 1. Is there a way to prove item 1, before item 2? I mean, not using item 2 in item 1?
 A: Use the sub index notation.
\begin{align}
\nu (v_1,\ldots, v_j,\ldots, v_n) 
=& 
\nu \left(\sum_{j_1=1}^{n} a_{1j_1}e_{j_1},\ldots, \sum_{j_i}^{n} a_{ij_i}e_{j_i},\ldots, \sum_{j_n}^{n} a_{nj_{n}}e_{j_{n}}\right)
\\
=&
\sum_{j_1=1}^{n} a_{1j_1} \cdot \nu \left(e_{j_1},\sum_{j_2}^{n} a_{2j_2}e_{j_2},\ldots, \sum_{j_i}^{n} a_{ij_i}e_j,\ldots, \sum_{j_n}^{n} a_{nj_{n}}e_{j_{n}}\right)
\\
=&
\sum_{j_1=1}^{n} a_{1j_1} \cdot \sum_{j_2=1}^{n} a_{2j_2}  \cdot \nu \left(e_{j_1},e_{j_2}, \sum_{j_2}^{n} a_{2j_2}e_{j_3},\ldots, \sum_{j_n}^{n} a_{nj_{n}}e_{j_{n}}\right)
\\
=&
\sum_{j_1=1}^{n}\sum_{j_2=1}^{n}  a_{1j_1} \cdot  a_{2j_2}  \cdot \nu \left(e_{j_1},e_{j_2}, \sum_{j_2}^{n} a_{2j_2}e_{j_3},\ldots, \sum_{j_n}^{n} a_{nj_{n}}e_{j_{n}}\right)
\\
=&
\\
\vdots& 
\\
=&
\sum_{j_1=1}^{n}\ldots \sum_{j_i=1}^{n}\ldots \sum_{j_n=1}^{n} (a_{ij_i} \cdot\ldots \cdot   a_{ij_i}\cdot \ldots  a_{nj_n}) \cdot \nu \left(e_{j_1},\ldots,  e_{j_i},\ldots, e_{j_{n}}\right)
\end{align}
Here is the subtlety to understand. Each of the sequences of length $n$ $j_1,j_2,\ldots,j_i,\ldots,j_n$ is a function $\{1,\ldots,i,\ldots, n\}\ni i\mapsto j_{i}\in \{1,\ldots,i,\ldots, n\} $. To see this, denote the sequence $j_1,j_2,\ldots,j_i,\ldots,j_n$ by $\tau$ and observe that $\tau(1)=j_1, \ldots, \tau(i)=j_i, \ldots, \tau(n)=j_n$.  Then, if $S^{n}$ is the set of all permutation $\sigma$ in $\{1,\ldots,i,\ldots, n\}$ such that $\sigma(1)=j_1,\ldots, \sigma(i)=j_i,\ldots, \sigma(n)=j_n$ we have
$$
\nu (v_1,\ldots, v_j,\ldots, v_n) = \sum_{\sigma\in S^{n}}(a_{ij_i} \cdot\ldots \cdot   a_{ij_i}\cdot \ldots  a_{nj_n}) \cdot \nu \left(e_{j_1},\ldots,  e_{j_i},\ldots, e_{j_{n}}\right)
$$
and
\begin{align}
\nu (v_1,\ldots, v_j,\ldots, v_n) 
=&
\sum_{\sigma\in S^{n}}(a_{1\sigma(1)} \cdot\ldots \cdot   a_{i\sigma(i)}\cdot \ldots  a_{n\sigma(n)}) \cdot \nu \left(e_{\sigma(1)},\ldots,  e_{\sigma(i)},\ldots, e_{\sigma(n)}\right)
\\
=&
\sum_{\sigma\in S^{n}}(a_{1\sigma(1)} \cdot\ldots \cdot   a_{i\sigma(i)}\cdot \ldots  a_{n\sigma(n)}) \cdot \mathop{\rm sng}(\sigma)
\\
=&
\det\big( a_{ij} \big)
\end{align}
A similar argument can be used for item 2.
