Let $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ be a differentiable map, given by $$f(x_{1},...,x_{n}) = (y_{1},...,y_{n}),$$ and $\omega = dy_{1}\wedge \cdots dy_{n}$. Show that $$f^{*}\omega = \det(df)dx_{1}\wedge \cdots \wedge dx_{n}.$$
My attempt.
$$\begin{eqnarray*} f^{*}\omega & = & f^{*}(dy_{1}\wedge \cdots \wedge dy_{n})\\ & = & f^{*}(dy_{1})\wedge \cdots \wedge f^{*}(dy_{n})\\ & = & d(f^{*}y_{1}) \wedge \cdots \wedge d(f^{*}dy_{n})\\ & = & d(y_{1}\circ f)\wedge \cdots \wedge d(y_{n}\circ f)\\ & = & df_{1}\wedge \cdots \wedge df_{n}\\ & = & \sum_{j=1}^{n}\frac{\partial f_{1}}{\partial x_{j}}dx_{j}\wedge \cdots \wedge \sum_{j=1}^{n}\frac{\partial f_{n}}{\partial x_{j}}dx_{j}, \end{eqnarray*}$$
but, $dx_{i} \wedge dx_{i} = 0$. So, $$\frac{n!}{n!(n-n!)} = 1$$ is the number of distinct combinations of $dx_{i}$. Let $\sigma$ be a permutation of $n$ elements, then
$$\begin{eqnarray*} f^{*}\omega & = & \sum_{j=1}^{n}\frac{\partial f_{1}}{\partial x_{j}}dx_{j}\wedge \cdots \wedge \sum_{j=1}^{n}\frac{\partial f_{n}}{\partial x_{j}}dx_{j}\\ & = & \sum \frac{\partial f_{1}}{\partial x_{\sigma(1)}}\cdots \frac{\partial f_{n}}{\partial x_{\sigma(n)}}dx_{\sigma(1)} \wedge \cdots \wedge dx_{\sigma(n)}. \end{eqnarray*}$$
Now, I know that changing $dx_{\sigma(1)} \wedge \cdots \wedge dx_{\sigma(n)}$ with $dx_{1} \wedge \cdots \wedge dx_{n}$ should be appear $\mathrm{sign}(\sigma)$ in the sum, but I cannot justify. Probably it is trivial, but can someone help me? I know that this question was asked here, but I didn't find this detail.