Since $rank(A)>rank(A^2)$, there is $k\in[[2, n]]$ and an invertible $P$ s.t. $P^{-1}AP=diag(J_k,U_{n-k})$ where $J_k$ is the nilpotent Jordan block of dimension $k$. Thus we choose $B=Pdiag({J_k}^{k-1},0_{n-k})P^{-1}$.

The result remain the same if we change the hypothesis with $\operatorname{rank} A^k \neq \operatorname{rank} A^p.$ , for some different natural k,p.

i cant understand why from $rank(A)>rank(A^2)$ we get there is $k\in[[2, n]]$ and an invertible $P$ s.t. $P^{-1}AP=diag(J_k,U_{n-k})$

Since $A$ is not invertible, in the Jordan form of $A$, there are nilpotent jordan blocks. If all these blocks were $[0]$-shaped, then $rank(A)=rank(A^2)$; then at least one block has dimension $\geq 2$.