The question is:

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

d y_1\wedge...\wedge d y_n=(detD \phi(x)) \cdotd x_1\wedge...\wedged 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 " \det A = \sum_{\sigma\in S_n}\operatorname{sign}(\sigma)\prod_{i=1}^na_{i \sigma(j)}"?