Prove that for a square matrix $A$ with $\mathrm{rank}(A^2) = \mathrm{rank}(A)$, we also have that $\mathrm{rank}(A^k) = \mathrm{rank}(A)$ for all $k\ge 2$.
I used the inequality of Frobenius to see that $$\mathrm{rank}(A^2) + \mathrm{rank}(A^2) \le \mathrm{rank}(A) +\mathrm{rank}(A^3).$$
Then utilizing the given equality we see that $\mathrm{rank}(A) \le \mathrm{rank}(A^3)$ which along with the inequality $$\mathrm{rank}(AB) \le \min(\mathrm{rank}(A),\mathrm{rank}(B))$$ would imply that $\mathrm{rank}(A) = \mathrm{rank}(A^3)$.
Is this proof valid? I understand I need to deal with the rest of the $k$ through the same process.
This is a valid proof. For another proof using linear transformations and their images see A square matrix $A$ is such that $\mathrm{rank}(A^{k}) = \mathrm{rank}(A^{k+1})$.