I'm reading about Axler Sheldon's [Down with Determinants](http://www.axler.net/DwD.pdf) (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.