  For an $m$-summable operator $A$ in a separable Hilbert space the higher
regularized Fredholm determinant $\det\nolimits_m(I+A)$ generalizes the
classical Fredholm determinant. Recently, Britz et al presented a proof of a
product formula \[
  \det\nolimits_m\bigl( (I+A)\cdot(I+B) \bigr)
  = \det\nolimits_m (I+A) \cdot \det\nolimits_m (I+B) \cdot
\exp\operatorname{Tr}\bigl({X_m(A,B)}\bigr), \] where $X_m(A,B)$ is an explicit
polynomial in $A,B$ with values in the trace class operators. If $m=1$ then
$X_1(A,B)=0$, hence the formula generalizes the classical determinant product
formula.
  One of the purposes of this note is to present two very simple alternative
proofs of the formula. The first proof is a priori analytic and makes use of
the fact that $z\mapsto \det\nolimits_m(I+zA)$ is holomorphic, while the second
proof is completely algebraic. The algebraic proof has, in our opinion, some
interesting aspects in its own about the trace and commutators.
  Secondly, we extend the above mentioned formula to several factors \[
  \det\nolimits_m\Bigl( \prod_{l=1}^r (I+A_l) \Bigr)
  =\left( \prod_{l=1}^r \det\nolimits_m (I+A_l) \right)
  \cdot \exp\operatorname{Tr}\bigl({X_{m,r}(A_r,\ldots,A_r)}\bigr). \] The
latter is more than just a straightforward generalization as we will gain more
insights into the combinatorics behind it. Also we will present an algebraized
version of the analytic proof in the language of formal power series. The
upshot is that the two identities are just combinatorial in nature.
