I'm attempting to prove the following:
Let $A$ be an operator on a finite-dimensional vector space over $\mathbb{C}$. Prove that, for each nonnegative integer $n$, $$\operatorname{dim}\operatorname{ker}A^{n+1}=\operatorname{dim}\operatorname{ker}A+\sum_{k=1}^n\operatorname{dim}(\operatorname{im}A^k\cap\operatorname{ker}A).$$
Certainly I'll need to invoke rank-nullity, and this is likely best done using induction, but I'm getting stuck on the details. More importantly though, I'm trying to get some kind of visual understanding of what's going on here. It's clear that $\operatorname{ker}A\subseteq\operatorname{ker}{A^2}\subseteq...\subseteq\operatorname{ker}A^n$, but I'm having a hard time grasping $\operatorname{im}A^k\cap\operatorname{ker}A$. Even in the first non-trivial case, it's not obvious to me why it should be true that $$\operatorname{dim}(\operatorname{ker}A^2)=\operatorname{dim}(\operatorname{ker}A)+\operatorname{dim}(\operatorname{im}A\cap\operatorname{ker}A).$$
As you suggested, the claim can be proved by induction. For $n=0$ there is nothing to show. For the induction step, let $B := A^n|_{\ker A^{n+1}}$. Then $\operatorname{im}(B) = \operatorname{im}(A^n)\cap\ker A$. This is easy to see. Hence, the rank-nullity formula for $B$ yields $$ \dim\ker A^{n+1} = \dim\ker B + \dim\operatorname{im}B = \dim\ker A^n + \dim(\operatorname{im}(A^n)\cap\ker A). $$