Normal Basis Theorem Proof

Letting $\equiv$ denote congruence modulo $f(x)$, we have $$ \det\left(\left(\sigma_i\sigma_kg(x)\right)_{ik}\right)^2 $$ $$ =\det\left(\left(\sum_j(\sigma_j\sigma_ig(x))(\sigma_j\sigma_kg(x))\right)_{ik}\right) $$ $$ =\det\left(\left(\sum_j\sigma_j(g_i(x)g_k(x))\right)_{ik}\right) $$ $$ \equiv\det\left(\left(\sum_j\sigma_j(\delta_{ik}g_i(x))\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\sum_j\sigma_j\sigma_ig(x)\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\sum_j\sigma_jg(x)\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\sum_jg_j(x)\right)_{ik}\right) $$ $$ =\det\left(\left(\delta_{ik}\right)_{ik}\right) $$ $$ =1. $$