Q:

What's the Jordan form of $J^2$?

$$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$
What's the Jordan form of $J^2$?
I know that it has two blocks and $2$ independent eigenvectors.  
So it could be 
$$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}
\text{ or } 
\begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$
But why can't the Jordan chain have the length of $4$ and $1$?

A:

You have characteristic polynomial $P(x)=x^5$ and minimal polynomial $m(x)=x^3$. This tells you there is a Jordan block for the eigenvalue $0$ of size $3$ (the multiplicity of the eigenvalue as a root of the minimal polynomial). This eliminates the $4$ and $1$ option you are worried about.
Indeed you have two options up to permutation of the blocks, either a $3$ block and $2$ blocks of size $1$, or a $3$ block and a $2$ block.

