[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*}
[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]
[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]
