How to prove $\det{AB}=\det{A}\det{B}$ with Leibniz formula in terms of Levi-Civita symbol and Einstein summation notation?

Question

Prove that $\det{AB}=\det{A}\det{B}$ with Leibniz formula in terms of Levi-Civita symbol and Einstein summation notation

Here is a similar question asked 6 years ago. The OP answered in the question that "expanding will give the answer which is then trivial by inspection", but I still can't understand why it is "trivial".

Answer 1

Let \begin{align} C=AB \end{align} then we have that \begin{align} c_{ij} = a_{ik}b_{kj}. \end{align} Next, notice that \begin{align} \det(C) =&\, \epsilon_{ijk}c_{1i}c_{2j}c_{3k}\\ =&\, \epsilon_{ijk}a_{1p}b_{pi}a_{2q}b_{qj}a_{3r}b_{rk}\\ =&\, a_{1p}a_{2q}a_{3r}\epsilon_{ijk}b_{pi}b_{qj}b_{rk}\\ =&\, a_{1p}a_{2q}a_{3r}\epsilon_{pqr}\det(B)\\ =&\, \det(A^T)\det(B) = \det(A)\det(B). \end{align}