[proofplan] We expand the Vandermonde product after permuting the entries and compare each factor with the corresponding factor in the original product. The permutation $\sigma$ sends ordered index pairs $i<j$ bijectively to unordered two-element subsets of $\{1,\dots,n\}$. A factor keeps its sign exactly when $\sigma(i)<\sigma(j)$ and changes sign exactly when $\sigma(i)>\sigma(j)$, so the total sign is $(-1)$ raised to the number of inversions of $\sigma$. By the definition of the signature through inversion parity, this total sign is $\operatorname{sgn}(\sigma)$. [/proofplan]

[step:Expand the permuted Vandermonde product over ordered index pairs] Let \begin{align*} I=\{(i,j)\in \{1,\dots,n\}^2: i<j\} \end{align*} denote the set of ordered pairs used in the Vandermonde product. By the definition of $\Delta$, \begin{align*} \Delta(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})=\prod_{(i,j)\in I}(\alpha_{\sigma(j)}-\alpha_{\sigma(i)}). \end{align*} The product is taken in the field $E$. Since $E$ is commutative, we may reorder finite products without changing their value. [/step]

[step:Record the sign contributed by each unordered pair]Define the set of inversions of $\sigma$ by \begin{align*} \operatorname{Inv}(\sigma)=\{(i,j)\in I:\sigma(i)>\sigma(j)\}. \end{align*} Let $\operatorname{inv}(\sigma)=|\operatorname{Inv}(\sigma)|$ denote the inversion number of $\sigma$. For each $(i,j)\in I$, let $a_{ij}=\min\{\sigma(i),\sigma(j)\}$ and $b_{ij}=\max\{\sigma(i),\sigma(j)\}$. Then $1\le a_{ij}<b_{ij}\le n$, and \begin{align*} \alpha_{\sigma(j)}-\alpha_{\sigma(i)}=(-1)^{\mathbb{1}_{\operatorname{Inv}(\sigma)}(i,j)}(\alpha_{b_{ij}}-\alpha_{a_{ij}}). \end{align*} Indeed, if $\sigma(i)<\sigma(j)$, then $a_{ij}=\sigma(i)$ and $b_{ij}=\sigma(j)$, so the factor is $\alpha_{b_{ij}}-\alpha_{a_{ij}}$. If $\sigma(i)>\sigma(j)$, then $a_{ij}=\sigma(j)$ and $b_{ij}=\sigma(i)$, so \begin{align*} \alpha_{\sigma(j)}-\alpha_{\sigma(i)}=\alpha_{a_{ij}}-\alpha_{b_{ij}}=-(\alpha_{b_{ij}}-\alpha_{a_{ij}}). \end{align*}[/step]

[guided]We need to compare the factor $\alpha_{\sigma(j)}-\alpha_{\sigma(i)}$ with the standard Vandermonde factor, whose smaller index must appear first as $\alpha_b-\alpha_a$ with $a<b$. For this reason, for each pair $(i,j)\in I$, we define \begin{align*} a_{ij}=\min\{\sigma(i),\sigma(j)\} \end{align*} and \begin{align*} b_{ij}=\max\{\sigma(i),\sigma(j)\}. \end{align*} These are elements of $\{1,\dots,n\}$ and satisfy $a_{ij}<b_{ij}$ because $\sigma$ is injective, so $\sigma(i)\ne \sigma(j)$ whenever $i\ne j$. There are exactly two cases. If $\sigma(i)<\sigma(j)$, then the permuted factor already has the Vandermonde orientation: \begin{align*} \alpha_{\sigma(j)}-\alpha_{\sigma(i)}=\alpha_{b_{ij}}-\alpha_{a_{ij}}. \end{align*} If $\sigma(i)>\sigma(j)$, then the order has been reversed, and the factor has the opposite sign: \begin{align*} \alpha_{\sigma(j)}-\alpha_{\sigma(i)}=\alpha_{a_{ij}}-\alpha_{b_{ij}}=-(\alpha_{b_{ij}}-\alpha_{a_{ij}}). \end{align*} The second case is precisely the condition that $(i,j)$ belongs to the inversion set $\operatorname{Inv}(\sigma)$. Thus every inversion contributes one factor of $-1$, and every non-inversion contributes no sign change: \begin{align*} \alpha_{\sigma(j)}-\alpha_{\sigma(i)}=(-1)^{\mathbb{1}_{\operatorname{Inv}(\sigma)}(i,j)}(\alpha_{b_{ij}}-\alpha_{a_{ij}}). \end{align*}[/guided]

[step:Reindex the unsigned factors by all two-element index pairs] Define \begin{align*} J=\{(a,b)\in \{1,\dots,n\}^2:a<b\}. \end{align*} The map $I\to J$ given by \begin{align*} (i,j)\mapsto (a_{ij},b_{ij}) \end{align*} is a bijection. It is injective because $\sigma$ is a bijection on $\{1,\dots,n\}$, so the unordered pair $\{\sigma(i),\sigma(j)\}$ determines the unordered pair $\{i,j\}$, and the condition $i<j$ then determines $(i,j)$. Since $I$ and $J$ are finite sets with the same cardinality, injectivity implies bijectivity. Therefore \begin{align*} \prod_{(i,j)\in I}(\alpha_{b_{ij}}-\alpha_{a_{ij}})=\prod_{(a,b)\in J}(\alpha_b-\alpha_a)=\Delta(\alpha_1,\dots,\alpha_n). \end{align*} [/step]

[step:Collect the inversion signs and identify them with the signature] Using the factor comparison and the reindexing above, we obtain \begin{align*} \Delta(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})=\prod_{(i,j)\in I}(-1)^{\mathbb{1}_{\operatorname{Inv}(\sigma)}(i,j)}\prod_{(i,j)\in I}(\alpha_{b_{ij}}-\alpha_{a_{ij}}). \end{align*} The first product contains one factor of $-1$ for each inversion and one factor of $1$ for each non-inversion, hence \begin{align*} \prod_{(i,j)\in I}(-1)^{\mathbb{1}_{\operatorname{Inv}(\sigma)}(i,j)}=(-1)^{\operatorname{inv}(\sigma)}. \end{align*} By the definition of the signature as the parity of the inversion number, \begin{align*} \operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}. \end{align*} Substituting this and the reindexed Vandermonde product gives \begin{align*} \Delta(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)})=\operatorname{sgn}(\sigma)\Delta(\alpha_1,\dots,\alpha_n). \end{align*} This is the desired identity. [/step]