Prove $\det(I + B) = 2(1 + tr(B)).$

Note that $\det B = 1$.

Note additionally that $B^{-1} = A(A^T)^{-1}$, so that $$ \operatorname{trace}(B^{-1}) = \operatorname{trace}(A(A^T)^{-1}) = \operatorname{trace}((A^T)^{-1}A) = \operatorname{trace}(B^T) = \operatorname{trace}(B) $$ Now, your polynomial can be written as $$ \det(I + B) = \det(B) + \det(B)\operatorname{trace}(B^{-1}) + \operatorname{trace}(B) + 1 $$ the conclusion follows.