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2 added 4 characters in body

$$A \otimes B = (A \circ I_n) \otimes (I_m \circ B) = (A \otimes I_m) \circ (I_n \otimes B)$$

where the second equation follows from functoriality of the tensor product. Here both $A \otimes I_m$ and $I_n \otimes B$ are square matrices of size $m n \times m n$. Since the determinant from such matrices to the scalar field is a monoid homomorphism, the determinant of the last expression is

$$\det(A \otimes I_m) \det(I_n \otimes B)$$

so we are left to determine the two determinants above. Since these are similar, we do the first. We may express an $m$-dimensional vector space $k^m$ as a direct sum of 1-dimensional vector spaces, so

$$A \otimes I_{k^m} = A \otimes (I_k \oplus \ldots \oplus I_k) = (A \otimes I_k) \oplus \ldots \oplus (A \otimes I_k)$$

because tensor products preserve direct sums. This is just $A \oplus \ldots \oplus A$. This matrix consists of $m$ blocks of $A$, so its determinant is $\det(A)^m$, and we are done.

$$A \otimes B = (A \circ I_n) \otimes (I_m \circ B) = (A \otimes I_m) \circ (I_n \otimes B)$$
where the second equation follows from functoriality of the tensor product. Here both $A \otimes I_m$ and $I_n \otimes B$ are square matrices of size $m \times n$. Since the determinant from such matrices to the scalar field is a monoid homomorphism, the determinant of the last expression is
$$\det(A \otimes I_m) \det(I_n \otimes B)$$
so we are left to determine the two determinants above. Since these are similar, we do the first. We may express an $m$-dimensional vector space $k^m$ as a direct sum of 1-dimensional vector spaces, so
$$A \otimes I_{k^m} = A \otimes (I_k \oplus \ldots \oplus I_k) = (A \otimes I_k) \oplus \ldots \oplus (A \otimes I_k)$$
because tensor products preserve direct sums. This is just $A \oplus \ldots \oplus A$. This matrix consists of $m$ blocks of $A$, so its determinant is $\det(A)^m$, and we are done.