An endomorphism $f: V \rightarrow V$ of an $F$-vector space is called nilpotent iff there exists $ \delta \in \mathbb N$ such that $f^\delta=0$. Suppose that $f
: V\rightarrow V$ is a nilpotent endomorphism of a finite dimensional vector space. Show that the vector space $V$ has an ordered basis $A$ such that the representing matrix $_A [f]_A$ of $f$ with respect to the basis has the form of an upper triangular matrix with only $0$s along the diagonal.
I've read proofs in the opposite order, but I'm not going to fall for that logical flaw.
So I've been trying to expand out the multiplication but that seems long winded, and I know that once I've proven it's upper triangular that I can use the Cayley-Hamilton theorem to show that the diagonal is zero. Any hints?
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