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?

$\textbf{Hint:}$ 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 $A^{n+k}(Ax)=0\implies A^{n}(Ax)=0\implies A^{n+1}(x)=0\implies A^{n}(x)=0$.
