Each linear operator has upper-triangle matrix for some basis? I'm reading about Axler Sheldon's Down with Determinants (page 8) and I read about a lemma I can't quite understand:

Lemma 6.1
Suppose $T $ is nilpotent. Then there is a basis of $V $ with respect
  to which the matrix of $T $ contains only $0$'s on and below the main
  diagonal.
Proof: First choose a basis of ker $T $. Then extend this to a basis of
  ker $T^2$. Then extend to a basis of ker $T^3$. Continue in this
  fashion, eventually getting a basis of $V $. The matrix of $T $ with
  respect to this basis clearly has the desired form.

Why does $T $ have the desired form? This is not clear for me. $V $ is a complex vector space.
Thank you for any help.
 A: It is simply because $$T(\ker(T^{k+1})\setminus\ker(T^k))\subset \ker (T^k)\setminus\ker(T^{k-1}).$$
So if Indeed if in this basis $\ker(T^{k-1})=\langle e_1,\dots,e_m \rangle$, $\ker(T^k)=\langle e_1,\dots,e_m,\dots ,e_n \rangle$  and $\ker(T^{k+1})=\langle e_1,\dots,e_n,\dots,e_{N} \rangle$ then $T$ must satisfy
$$T e_k=\sum_{i=m+1}^n\lambda_ie_i, k\in n+1,\dots,N,$$
i.e. in this basis $T$ is upper triangular.
A: Let's try it on an example. If $\dim V=4$, $T^3=0$ and if $(v_1,v_2)$ is a basis of $\text{Ker }T$, $(v_1,v_2,v_3)$ a basis of $\text{Ker }T^2$ and $(v_1,\dots,v_4)$ a basis of $\text{Ker }T^3$ then, note that for all $k\in\Bbb N$, we have 
$$T(\text{Ker }T^{k+1})\subseteq \text{Ker }T^k$$
This is because if $v\in\text{Ker }T^{k+1}$, then $T^k(T(v))=T^{k+1}(v)=0$ i.e. $T(v)\in \text{Ker }T^k$. 
Hence $T(v_1)=T(v_2)=0$, $T(v_3)=a v_1+b v_2\in \text{Ker } T$ for some reals $a,b$ and $T(v_4)=c v_1+d v_2+e v_3\in \text{Ker } T^2$ for some $c,d,e$. 
Thus the matrix of $T$ in this basis is
$$\begin{bmatrix}0&0&a&c\\ 0&0&b&d\\ 0&0&0&e\\ 0&0&0&0\\ \end{bmatrix}$$
A: The crucial thing is that $\ker T\subset \ker T^2\subset\ldots$. Let's take the first step only, i.e. assume $T^2=0$. Let $\{v_i\}$ be a basis of $\ker T$ and $
\{w_j\}$ an extension to a basis of $\ker T^2$ ($=\mathbb{R}^n$ as $T^2=0$). Hence
$$
[v_1,v_2,\ldots,v_k,w_1,w_2,\ldots,w_m]=[V\ W]=S
$$
is an invertible matrix. We have by definition $Tv_i=0$ and $T^2w_j=T(Tw_j)=0$, hence, $Tw_j\in\ker T$ and can be written in the basis $\{v_i\}$ as $Tw_j=\sum \lambda_{ij}v_i$. In the matrix notations: $TV=0$ and $TW=V\Lambda$, i.e.
$$
T[V\ W]=[V\ W]\begin{bmatrix}0 & \Lambda\\0 & 0\end{bmatrix}\qquad\Leftrightarrow\qquad S^{-1}TS=\begin{bmatrix}0 & \Lambda\\0 & 0\end{bmatrix}\text{ (upper triangular).}
$$
