Leibniz formula for determinants

Statement

Let $K$ be a commutative ring, let $n \ge 1$, and let $A = (a_{ij})$ be an $n \times n$ matrix over $K$. Let $S_n$ be the group of permutations of $\{1, \dots, n\}$, and let $\sgn(\sigma) \in \{1, -1\}$ be the sign of $\sigma$. Then the determinant of $A$ is $$\det A = \sum_{\sigma \in S_n} \sgn(\sigma) \prod_{i=1}^{n} a_{i, \sigma(i)}.$$