In $\mathbb R^n$ consider the differential form $\nu$ satisfying $\nu(e_1, \ldots, e_n)=1$. For every $i=1, \ldots, n$ consider the vector $\displaystyle v_i=\sum_{j=1}^n a_{ij} e_j$. How to show $\nu(v_1, \ldots, v_n)=\det(a_{ij})$? I can see it is true, but I can't argue why that is true..
I guess I should use the relation $$f^*\omega=\det(df)dx_1\wedge \ldots \wedge dx_n,$$ where $f:\mathbb R^n\longrightarrow \mathbb R^n$, $f=(f_1, \ldots, f_n)$, $y_i=f(x_1, \ldots, x_n)$ and $\omega=dy_1\wedge \ldots \wedge dy_n$. Using this, given $v_i=\sum_{j=1}^n a_{ij} e_j$, I defined $f:\mathbb R^n\longrightarrow \mathbb R^n$, $f=(f_1, \ldots, f_n)$, setting $$f_i(x_1, \ldots, x_n)=\sum_{j=1}^n a_{ij} x_j.$$
Then $\partial_j f_i=a_{ij}$ hence $$(f^*\omega)(e_1, \ldots, e_n)=\det(a_{ij})dx_1\wedge \ldots \wedge dx_n(e_1, \ldots, e_n)=\det(a_{ij}).$$ So the problem amounts showing $(f^*\omega)(e_1, \ldots, e_n)=\nu(v_1, \ldots, v_n)$, but still I can't do that.
Can anyone give me some hint?
Obs:
(i) I still don't know $\displaystyle\det(a_{ij})=\sum_{\sigma} \textrm{sign}(\sigma) a_{1\sigma(1)}\cdots a_{n\sigma(n)}$.
(ii) $(f^*\omega)(v_1, \ldots, v_n)=\omega(df(v_1), \ldots, df(v_n))$ (in simplified notation, i.e., ommiting the point the form came from).