Matrix of a Nilpotent Operator Proof

I'm trying to understand this one proof about the matrix of a nilpotent operator and I'm stuck on this one part. Here is the theorem I'm trying to prove and how the proof goes:

Suppose N is a nilpotent operator on a vector space V. Then there is a basis of V with respect to which the matrix of N has the form:

$\displaystyle

\[

\begin{bmatrix}

0 & & * \\

& \ddots & \\

0 & & 0

\end{bmatrix}

\]$

where all entries on and under the diagonal are 0 and the asterisk indicates the other entries.

The proof starts by creating a basis for V. First a basis for $\displaystyle null(N)$ is chosen and then extended to a basis of $\displaystyle null(N^2)$. Then our basis is extended to a basis for $\displaystyle null(N^3)$ and so on until $\displaystyle null(N^{\dim{V}})=V$. Now we have a basis of V.

The text then says:

"Now let’s think about the matrix of N with respect to this basis. The

first column, and perhaps additional columns at the beginning, consists

of all 0’s because the corresponding basis vectors are in $\displaystyle null(N)$. The

next set of columns comes from basis vectors in $\displaystyle null(N^2)$. Applying N

to any such vector, we get a vector in $\displaystyle null(N)$; in other words, we get a

vector that is a linear combination of the previous basis vectors. Thus

all nonzero entries in these columns must lie above the diagonal. The

next set of columns come from basis vectors in $\displaystyle null(N^3)$. Applying N

to any such vector, we get a vector in $\displaystyle null(N^2)$; in other words, we get a

vector that is a linear combination of the previous basis vectors. Thus,

once again, all nonzero entries in these columns must lie above the

diagonal. Continue in this fashion to complete the proof."

I understand how the entries in the first set of columns are all 0. However, I do not understand how the second set of columns are determined.