Here is the lemma: If $N \in L(V )$ is nilpotent, then there exist vectors $v_1,\dots,v_n\in V$ and nonnegative integers $m_1,\dots,m_n$ such that
(a) $(N^{m_1}(v_1),\ldots,N(v_1),v1; \ldots ;N^{m_n}(v_n),\ldots N(v_n),v_n)$ is a basis of $V$ and
(b) $N^{m_1+1}(v_1)=\ldots =N^{m_n+1}(v_n)=0$
(The proof is by induction on $\dim(V)$. I need to show the base case($\dim(V=1$)). All the proofs I have seen say this is trivial. However, I need to show it precisely, and in detail) Help?
Here is a proof of the base case: Assume that $\dim V = 1$. This implies that every linear operator on $V$ is multiplication by a scalar. Thus the only nilpotent operator on $V$ is the $0$ operator. With that information, we have $N = 0$ and we can take $n = 1$, $v_1$ to be any nonzero vector in $V$, and $m_1 = 0$.