Determinant/Leibniz formula/Fact

Leibniz formula for the determinant

For the determinant of an ${}n\times n$-matrix

${}M={\left(a_{ij}\right)}_{ij}\,,$

the formula

${}\det M=\sum _{\pi \in S_{n}}\operatorname {sgn} (\pi )a_{1\pi (1)}\cdots a_{n\pi (n)}\,$

holds.

Proof, Write another proof