Show that there exists $n\in\mathbb{Z}^+$ such that $\ker (S^n)=\ker (S^{n+1})$

Question

Any help with the following question...

Let $S: \textsf{V} \to \textsf{V}$ be a linear operator. Show that if $\textsf{V}$ is finite-dimensional, there exists $n\in\mathbb{Z}^+$ such that $\ker (S^n)=\ker (S^{n+1})$. In this case, $\ker (S^k) \nsubseteq \ker (S^{k+1})$ for all $k<n$.

Proof a similar result for $\text{im} (S^n)$.

Answer 1

Observe that

$$v\in\ker S\implies S^2v=S(Sv)=S(0)=0\implies \ker S\subset \ker S^2$$

So begin constructing an ascending series $\;\ker S\subset\ker S^2\subset\ker S^3\subset\ldots\;$, and then use finite dimensionality.