Proof that the determinant of a transpose equals the determinant

Proof

1. Let $A = (a_{ij})$. By the definition of the transpose, $(A^T)_{ij} = a_{ji}$.
2. By the Leibniz formula applied to $A^T$, $\det A^T = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^{n} a_{\sigma(i), i}$.
3. Fix $\sigma$. As $i$ runs over $1, \dots, n$, the index $k = \sigma(i)$ also runs over $1, \dots, n$, and $i = \sigma^{-1}(k)$.
4. Since $K$ is commutative, we may reorder the product. So $\prod_{i=1}^{n} a_{\sigma(i), i} = \prod_{k=1}^{n} a_{k, \sigma^{-1}(k)}$.
5. If $n \ge 2$, the sign is a homomorphism. So $\sgn(\sigma)\sgn(\sigma^{-1}) = \sgn(\sigma \sigma^{-1}) = 1$.
6. Both signs lie in $\set{1, -1}$, so step 5 gives $\sgn(\sigma^{-1}) = \sgn(\sigma)$. If $n = 1$, then $\sigma$ is the identity and this is clear.
7. By steps 2, 4 and 6, $\det A^T = \sum_{\sigma \in S_n} \sgn(\sigma^{-1}) \prod_{k=1}^{n} a_{k, \sigma^{-1}(k)}$.
8. The map $\sigma \mapsto \sigma^{-1}$ is a bijection of the symmetric group $S_n$ onto itself. So we may write $\rho = \sigma^{-1}$ and sum over $\rho \in S_n$.
9. This gives $\det A^T = \sum_{\rho \in S_n} \sgn(\rho) \prod_{k=1}^{n} a_{k, \rho(k)} = \det A$, by the Leibniz formula for $A$.