Determinant function/Uniquely determined/Fact

The universal property of the determinant

Let ${}K$ be a field and ${}n\in \mathbb {N}$.

Then there exists exactly one

determinant function

$\triangle \colon \operatorname {Mat} _{n}(K)=(K^{n})^{n}\longrightarrow K$

fulfilling

${}\triangle {\left(e_{1},\,e_{2},\ldots ,e_{n}\right)}=1\,,$

where ${}e_{i}$ denote the standard vectors, namely the

determinant.

Proof, Write another proof