The question is:

Let $\displaystyle \phi: \mathbb{R}^n\rightarrow\mathbb{R}^n$ be a $\displaystyle C^1$ map and let $\displaystyle y=\phi(x)$ be the change of variables. Show that

d$\displaystyle y_1\wedge...\wedge $d$\displaystyle y_n$=(detD$\displaystyle \phi(x)$)$\displaystyle \cdot$d$\displaystyle x_1\wedge...\wedge$d$\displaystyle x_n$.

Take a look at here and the answer given by Michael Albanese:

differential geometry - wedge product and change of variables - Mathematics Stack Exchange

My question is can we prove it without using the fact "$\displaystyle \det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i \sigma(j)}$"?