The Vandermonde product transforms by the sign of the root permutation

Swapping two entries of the ordered root list reverses the factor belonging to that pair, while the remaining affected factors exchange in pairs; hence a transposition multiplies $\delta$ by $-1$.

Taking the one-factor factorisation $\tau =\tau$ in [L3] gives $(-1{)}^{\mathrm{inv}(\tau )}=-1$, so $\mathrm{sgn}(\tau )=-1$ for every transposition $\tau$ by [F1].

By [L2] write $\sigma ={\tau }_1\cdots {\tau }_r$ as a product of transpositions, and apply step 1.1 once for each factor: each application multiplies the current Vandermonde product by $-1$, so $\sigma (\delta )=(-1{)}^r\delta$.

By [L1] and step 1.2, $\mathrm{sgn}(\sigma )=\mathrm{sgn}({\tau }_1)\cdots \mathrm{sgn}({\tau }_r)=(-1{)}^r$, so step 2.1 gives $\sigma (\delta )=\mathrm{sgn}(\sigma )\delta$. For $n=0$ or $n=1$ the factorisation is empty by [L2], the product defining $\delta$ is likewise empty and equals $1$, and the sign is $1$, so the identity holds there as $1=1$.

Squaring the identity of step 3.1 removes the sign, so $\sigma ({\delta }^2)={\delta }^2$ for every $\sigma$. A root equal to zero creates no exception; separability ensures distinct roots and hence $\delta \mathrm{\ne }0$.