Determinant of the transpose

Leibniz formula.
$$\det A = \sum_{\sigma\in S_n}\varepsilon(\sigma)\prod_{i=1}^n a_{\sigma(i),i}.$$

Transpose. The entry $(i,j)$ of $A^T$ is $a_{ji}$. So
$$\det(A^T) = \sum_{\sigma\in S_n}\varepsilon(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}.$$
Since $\prod_{i=1}^n a_{i,\sigma(i)}=\prod_{j=1}^n a_{\sigma^{-1}(j),j}$ (reindexing $j=\sigma(i)$), and $\varepsilon(\sigma^{-1})=\varepsilon(\sigma)$ (since $\varepsilon$ is a group homomorphism and $\sigma\sigma^{-1}=e$), we can rewrite:
$$\det(A^T) = \sum_{\sigma}\varepsilon(\sigma^{-1})\prod_j a_{\sigma^{-1}(j),j} = \sum_{\tau}\varepsilon(\tau)\prod_j a_{\tau(j),j} = \det A,$$
where the last change of variables is $\tau=\sigma^{-1}$.

Conclusion. $\det(A^T)=\det A$. In particular, properties of columns hold for rows too.