Nilpotent linear map $x$ induces another nilpotent linear map $ \bar{x} : V/ U \rightarrow V/ U $ that has a basis that is strictly upper triangular. I'm having trouble proving an exercise (Exercise 6.1 in Erdmann and Wildon's book Introduction to Lie Algebras). The exercise is used to help prove a version of Engel's Theorem. 
It states: 
Let $V$ be an $n$-dimensional vector space where $ n \geq 1 $ and let $ x : V \rightarrow V $ be a nilpotent linear map. 
(i): Show that there is a non-zero vector $v \in V $ such that $ xv = 0 $. (this part I've done). 
(ii): Let $U  = \text{Span} \{ v \} . $ Show that $ x $ induces a nilpotent linear transformation $ \bar{x} : V/U \rightarrow V/U$. (this part again I've showed and proved it's well defined). 
This is the part of the question I'm struggling with: 
By induction we know that there is a basis $ \{ v_1 + U , ..., v_{n-1} +U \} $ of $ V/ U $ in which $ \bar{x} $ has a strictly upper triangular matrix. Prove that $ \{ v_1,...,v_{n-1} \} $ is a basis of $ V$ and that matrix of $x$ is strictly upper triangular w.r.t this basis. 
I'm having an issue with the induction and the proof from there onwards. 
I don't really know where to start, for my $ \bar{x} $ I have $ \bar{x} = x + U $. Any help in the right direction would be appreciated, don't know if I'm missing something obvious. 
 A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Write
$$
V_{i} = \Span{ v_{i}, \dots, v_{n-1}, v}.
$$
You know that $\bar x$ maps $V_{i}/U$ to $V_{i+1}/U$ for $i = 1, \dots, n-1$, where $V_{n} = U$. Thus if $v \in V_{i}$, for $i = 1, \dots, n-1$, we have
$$
x v + U = \bar x (v + U) \in V_{i+1}/U,
$$
so that for $v \in V_{i}$ we have $x v \in V_{i+1}$. And then $x V_{n} = x U = \Set{0}$.
A: 
By induction we know that there is a basis $ \{ v_1 + U , ..., v_{n-1} +U \} $ of $ V/ U $ in which $ \bar{x} $ has a strictly upper triangular matrix.

Let's look at the very first step.  Suppose we extend to $v$ to a basis (an arbitrary basis) $\mathcal B = \{v,v_1,\dots,v_{n-1}\}$. We note that the matrix of $x$ with respect to $\mathcal B$ has the form
$$
[x]_{\mathcal B} = 
\left[
\begin{array}{c|ccc}
0 & *\\
\hline
0 & [\bar x]_{\bar{\mathcal B}} 
\end{array}
\right]
$$
Where $\bar {\mathcal B} = \{v_1 + U, \cdots, v_{n-1} + U\}$.  However, we know that $\bar x$ is itself nilpotent.  It follows that we can modify our choice of $v_1,\dots,v_{n-1}$ (so that $v_1 + U \in \ker \bar x$) to get
$$
[\bar x]_{\bar{\mathcal B}} = \left[
\begin{array}{c|ccc}
0 & *\\
\hline
0 & [\bar{\bar x}]_{\bar{\bar{\mathcal B}}} 
\end{array}
\right]
$$
With respect to this choice of $v_i$, the original matrix has the form
$$
[x]_{\mathcal B} = \pmatrix{0&*&*\\0&0&*\\0&0&[\bar{\bar x}]_{\bar{\bar B}}}
$$
The pattern continues.
