Does the Lie-Trotter product generalize to more than two matrices?

Question

Does the Lie-Trotter product formula generalize to more than two matrices? I.e.,

$e^{A+B+C} = \lim_{k \to \infty} \left( e^{\frac{A}k} e^{\frac{B}k} e^{\frac{C}k} \right)^k$ ?

Answer 1

The proof is from Theorem 2.11, Lie groups, Lie algebras, and representations, by Hall, 2015.

Proof: $e^{A/n}e^{B/n}=I+(A+B)/n+O(1/n^2)$, so $D_n:=\log(e^{A/n}e^{B/n})=(A+B)/n+O(1/n^2)$. Then $e^{A/n}e^{B/n}e^{C/n}=e^{D_n}e^{C/n}=e^{D_n+C/n+O(1/n^2)}$. Therefore, $(e^{A/n}e^{B/n}e^{C/n})^n=e^{nD_n+C+O(1/n)}=e^{A+B+C+O(1/n)}$.