Nilpotent endomorphism/Trigonalizable/Fact

Let ${}K$ be a field and let ${}V$ denote a finite-dimensional ${}K$-vector space. Let

$\varphi \colon V\longrightarrow V$

be a nilpotent linear mapping.

Then ${}\varphi$ is trigonalizable.

There exists a basis such that ${}\varphi$ is described, with respect to this basis, by an upper triangular matrix, in which all diagonal entries are ${}0$.

Proof, Write another proof