ker$(A^n) =$ ker$(A^m), \forall m > n$ 
If $A$ is a square matrix and ker$(A^n) =$ ker$(A^{n+1})$, then
  ker$(A^n) =$ ker$(A^m), \forall m > n$.

I'm trying to prove that this is correct, but I'm having trouble figuring out what the relation between the ker$(A^n)$ and ker$(A^{n+1})$ is precisely. I know "ker" represents all solutions to the homogenous equation, but I'm not sure how to think about something like ker$(A^n)$ where $A$ is raised to a power.
I can think of something like ker$(A_1\times ... \times A_n)$ = ker$(A_1\times ... \times A_{n+1})$, but is there some way I can transform this equation to make it easier to understand how to prove the above?
 A: You can think of the problem in the following way:

If we know that $A^{n+1}x = A^n \cdot Ax = 0 \implies A^n x = 0$ (for all $x$), show that
  $A^m x = 0 \implies A^n x = 0$ for all $m>n$ (for all $x$).

In fact, it's enough to show the following:

If we know that $A^{n+1}x = A^n \cdot Ax = 0 \implies A^n x = 0$ (for all $x$), show that
  $A^m x = 0 \implies A^{m-1} x = 0$ for all $m>n$ (for all $x$).

Now, note that
$$
A^m x = A^{n+1}(A^{m-n-1} x) = 0
$$
and apply the hypothesis.
A: $\textbf{Hint:}$ We want to prove that if $\ker(A^n)=\ker(A^{n+1})$, then $\ker(A^n)=\ker(A^{n+k})$ for all $k\in\mathbb{N}$.
We know this is true for $k=1$ by assumption, so assume that $\ker(A^n)=\ker(A^{n+k})$ for some $k\in\mathbb{N}$.
If $A^{n+k+1}(x)=0,$ then $A^{n+k}(Ax)=0\implies A^{n}(Ax)=0\implies A^{n+1}(x)=0\implies A^{n}(x)=0$.
A: We allways have that $\ker A^n\subseteq \ker A^{n+1}$ since multiplying by $A$ is just composing the transformation of $A^n$ with that of $A$ and $0$ always maps to $0$. So with induction you can prove $\ker A^n\subseteq\ker A^n$ if $m\geq n$ 
We shall now prove if $\ker A^{n+1}\subseteq\ker A^{n+1}$ then $\ker A^{n+2}\subseteq\ker A^{n+1}$. Suppose $A^{n+2}(v)=0$ then $A(v)\in \ker A^{n+1}=\ker A^n$ since $A(v)\in \ker A^n$ we have $A^{n+1}(v)=0$ so $A\in \ker A^{n+1}$.
As before induction proves $\ker A^m\subseteq \ker A^n$ if $m\geq n$
