If $\nu(e_{1},\dots,e_{n}) = 1$ and $v_{1} = \sum a_{ij}e_{j}$ then $\nu(v_{1},\dots,v_{n}) = det(a_{ij})$

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this is Do Carmo's "Diferential Forms and Applications" book problem, specifically problem $9$, in chapter $1$. This seems to be a problem to define the volume element of $\mathbb{R}^{n}$.

Let $\nu$ be a $n-$form in $\mathbb{R}^{n}$, with $\nu(e_{1},\dots,e_{n}) = 1$. If $v_{i} = \sum a_{ij}e_{j}$ then $\nu(v_{1},\dots,v_{n}) = det(a_{ij})$.

I've tried to use the "pullback" formula, proved here, but It didn't worked well. What should I do? Which steps are needed to prove the question? Thanks in advance!