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?

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.
