If $T: V \to V $ is an operator with $\mathrm{rank}(T)=n-1$, then there exists a vector $v \in V$ such that $T^{n-1}(v) \ne 0$ 
Let $V$ be an $n$-dimensional vector space over a field $F$ and $T: V \to V$ an operator. If $\mathrm{rank}(T)=n-1,$ prove that $T^{n-1}$ is not equal to the $0$ operator on $V$.

We need to show that there exists a vector $v \in V$ such that $T^{n-1}(v) \ne 0$. Since $\mathrm{rank}(T) = n-1$ and $\dim V = n$, we have $\dim \ker(T) = n -(n-1)=1$ by the rank-nullity theorem. Suppose $(w)$ is a basis of $\ker(T)$ and extend it to a basis $B_V = (v_1,\ldots,v_n)$ where $w = v_j$ for some $j \in \{1,\ldots,n\}$. Apply $T$ to a vector $v$ in the basis of $V$ but not contained in $\mathrm{span}(w)$. Then $T(v) \ne T(\alpha w) = \alpha T(w)$ for every $\alpha \in F$. Thus $T^{n-1}(v) \ne \alpha T^{n-1}(w) = \alpha T^{n-2}(T(w)) = \alpha T^{n-2}(0)=0 \Rightarrow T^{n-1}(v) \ne 0$.
Question 1: Is this correct? I am really struggling to prove this. 
Question 2: In the case where $V$ is a $1$-dimensional vector space, wouldn't  $T^{n-1}(v)$ be the $0$ operator on $V$? We have $\ker(T) \subset V$ and if $\mathrm{rank} = n -1 = 1 - 1 = 0$ then $\dim V = 0 + \mathrm{nullity}(T) = \dim \ker(T)$ implying that $V= \ker(T)$ and so $T(v)=0$ for every $v \in V$. Surely this implies that $T^{n-1}(v)$ is the $0$ operator? 

Question 1: No, it's not correct.  You can't conclude that $T^{n-1}(v)\neq \alpha T^{n-1}(w)$ just because $T(v)\neq \alpha T(w)$.  If you're still stuck on the problem, I suggest trying to prove that $\dim\ker(T^k)\leq k$ for all $k$ by induction on $k$.
Question 2: When $n=1$, the hypotheses do imply $T=0$.  However, $T^{n-1}$ is not $0$, because $n-1=0$, and any operator (even $0$) raised to the $0$th power is defined to be the identity operator.
