Suppose that $A$ is an $n\times n$ nilpotent matrix.
Let $x_1,\ldots,x_m$ be eigenvectors of $A$ forming a basis for the eigenspace of $A$.
For $j=1,\ldots,m$, form a Jordan chain $$C\left(x_j\right)=\left\{v_{j,1},\ldots,v_{j,k_j}\right\}$$ by "going backward", that is, start from the eigenvector $v_{j,1}=x_j$ and iteratively solve $$Av_{j,k}=v_{j,k-1}$$ while the system has a solution (until you find a system that has no solution).
Several sources I found on the web, as well as an old textbook, seem to suggest that $$C=C\left(x_1\right) \cup \ldots \cup C\left(x_m\right)$$ is a basis for the generalized eigenspace of $A$, that is, for the null space of $A^n$, without offering a formal proof that the algorithm works. How would you prove that? Any reference would be highly appreciated.
This approach does not work. Consider the matrix $$ A=\pmatrix{0&0&1\\0&0&0\\0&0&0} $$ with eigenvectors $\pmatrix{1\\1\\0}$ and $\pmatrix{1\\-1\\0}$. None of them is in the range of $A$, only their sum is.
Instead one has to work from the other direction: Assume $A^dv=0$ but $A^{d-1}v\ne 0$. Then $(A^{d-1}v \dots Av,v)$ are linear independent and part of the Jordan basis. A reference for this approach is the book by Axler.