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Split $f=f_{-}+a_0+f_{+}$ into strictly negative and strictly positive Fourier parts, and let $A=T(f_{-})$, $B=T(f_{+})$. The Toeplitz multiplication rule is $T(g)T(h)=T(gh)$ when $g$ is coanalytic or $h$ is analytic. It follows directly because the discarded negative Fourier part cannot return to the Hardy space in those cases. Thus exponentiation within either subalgebra is exact, andThe constant terms cancel. If $S=T(z)$ is the unilateral shift, $A={\sum }_{m>0}{a}_{-m}(S^{\ast }{)}^m$ and $B={\sum }_{n>0}{a}_n{S}^n$. The commutator $[(S^{\ast }{)}^m,S^n]$ is finite rank, with trace $n$ when $m=n$ and zero otherwise. For equal powers it is the projection onto the first $n$ basis vectors; for unequal powers its matrix has no diagonal entries. Its trace norm is bounded by $2\min (m,n)$. Smoothness of $f$ makes the weighted double coefficient sum finite, so the commutator series converges in trace norm andApplying part (c) proves the Toeplitz exponential determinant identityIn particular the product is an invertible identity-plus-trace-class operator, so the determinant in the formula is well-defined.