Show that rank($A^{n+1}$) = rank($A^n$) Suppose $A$ is a $n \times n$ matrix i.e. $A \in \mathbb{C}^{n \times n}$, prove that rank($A^{n+1}$) = rank($A^n$). In other words, I need to prove that their range spaces or null spaces are equal. If it helps, $A$ is a singular matrix.
Note that, I don't want to use Jordan blocks to prove this. Is it possible to prove this without using Jordan form? I can use Schur's triangularization theorem. Also, it's not known if A is diagonalizable.
$\newcommand{\rg}{\operatorname{range}}$
Obviously for every $m$, $\rg A^{m+1}\subset\rg A^{m}$, so if $d_m=\dim\rg A^m$, $d_{m+1}\le d_m$. If $d_{m+1}=d_m$ for some $m$, then $\rg A^{m+1}=\rg A^{m}$ and therefore $\rg A^m=\rg A^{m+1}=\rg A^{m+2}=\dotsb{}$. That is, the sequence $d_0,d_1,\dots$ becomes constant once it stops descending.

Because $d_0= n$, the sequence must stop descending within $n$ terms.
Edit: For the problem you phrased in the comment, $\rg A^{m+1}=\{AA^{m}y:y\in \mathbb C^n\}=\{Ax:x=A^my\in\rg A^m\}=\{Ax:x\in\rg A^m\}$,
Therefore  $\rg A^{m}=\rg A^{m+1}\implies$
$\rg A^{m+1}=\{Ax:x\in\rg A^m\}=\{Ax:x\in\rg A^{m+1}\}=\rg A^{m+2}$.
Every matrix corresponds to a linear map. Suppose that $A=\mathcal{M}(T)$, where $T\in \mathcal{L}(V)$ and $dim (V)=n$. Using Theorem 8.4 in Axler,
$$Null(T^n)=Null(T^{n+1}).$$
According to the Fundamental Theorem of Linear Maps,
$$n=dim Null(T^n)+ dim Range(T^n)$$
$$n=dim Null(T^{n+1})+dim Range(T^{n+1})$$
$dim Range(T^n)=dim Range(T^{n+1})$ implies that $rank(A^n)=rank(A^{n+1})$.