The product formula for regularized Fredholm determinants

For trace class operators $A, B \in \mathcal{B}_1(\mathcal{H})$ ($\mathcal{H}$ a complex, separable Hilbert space), the product formula for Fredholm determinants holds in the familiar form $${\det}_{\mathcal{H}} ((I_{\mathcal{H}} - A) (I_{\mathcal{H}} - B)) = {\det}_{\mathcal{H}} (I_{\mathcal{H}} - A) {\det}_{\mathcal{H}} (I_{\mathcal{H}} - B).$$ When trace class operators are replaced by Hilbert--Schmidt operators $A, B \in \mathcal{B}_2(\mathcal{H})$ and the Fredholm determinant ${\det}_{\mathcal{H}}(I_{\mathcal{H}} - A)$, $A \in \mathcal{B}_1(\mathcal{H})$, by the 2nd regularized Fredholm determinant ${\det}_{\mathcal{H},2}(I_{\mathcal{H}} - A) = {\det}_{\mathcal{H}} ((I_{\mathcal{H}} - A) \exp(A))$, $A \in \mathcal{B}_2(\mathcal{H})$, the product formula must be replaced by $${\det}_{\mathcal{H},2} ((I_{\mathcal{H}} - A) (I_{\mathcal{H}} - B)) = {\det}_{\mathcal{H},2} (I_{\mathcal{H}} - A) {\det}_{\mathcal{H},2} (I_{\mathcal{H}} - B) \exp(- {\rm tr}(AB)).$$ The product formula for the case of higher regularized Fredholm determinants ${\det}_{\mathcal{H},k}(I_{\mathcal{H}} - A)$, $A \in \mathcal{B}_k(\mathcal{H})$, $k \in \mathbb{N}$, $k \geq 2$, does not seem to be easily accessible and hence this note aims at filling this gap in the literature.