 Signed Jones evaluation at minus one

Use the Conway-normalized Alexander polynomial. Choosing $t^{1/2}=i$ in the Jones polynomial skein relation gives $V_{L_{+}}(-1)-V_{L_{-}}(-1)=-2i{V}_{L_0}(-1)$. The quantities $(-1{)}^{\ell -1}{\nabla }_L(2i)$ obey the same relation, because smoothing changes the component count by one. Their unlink values agree, so crossing-change induction proves equality for every link. For a knot, the sign is one and ${\nabla }_K(2i)={\Delta }_K(-1)$. This proves the signed identity, not only equality of knot determinants.