= Solution
For the root-discriminant convention in the requested identity, define
$$
\Delta_f=\prod_{i<j}(\alpha_i-\alpha_j)^2.
$$
The <Vandermonde determinant> is $\det V(\alpha)=\prod_{i<j}(\alpha_j-\alpha_i)$. Squaring removes the signs, proving
$$
\boxed{\Delta_f=(\det V(\alpha))^2}.
$$
It is nonzero because the roots are distinct. It lies in $F$, since its root expression is fixed by every <Galois automorphism> and $K^G=F$.
If $f$ is not monic and has leading coefficient $c$, the usual <polynomial discriminant> is $c^{2n-2}\prod_{i<j}(\alpha_i-\alpha_j)^2$. Thus the printed Vandermonde equality uses the discriminant of the monic normalization $f/c$. The extra factor is itself a nonzero square in $F$, so it does not alter the square-class criterion in the next part.
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