If $\ker f^k = \ker f^{k+1}$, then $\ker f^k = \ker f^{k+m}$ for all $m\in\mathbb N$ If $f: V\rightarrow V$ is a linear transformation, with $V$ being a finite-dimensional vector space over $\mathbb F$, such that $\ker f^k = \ker f^{k+1}$, then  $\ker f^k = \ker f^{k+m}$ for all $m\in\mathbb N$.
We prove by induction on $m$. If $m =1$, this is clearly true.
Now, assume it is true for $m=n$ and we show that it is true for $m=n+1$. Thus we need to show that $\ker f^k = \ker f^{k+n+1}$. 
Is it correct to say that since $\ker f^k = \ker f^{k+1}$ for any $k\in\mathbb N$, then $\ker f^{k+n}$ = $\ker f^{k+n+1}$, hence $\ker f^k = \ker f^{k+n+1}$?

Here's the inductive step: 
Suppose $\ker f^{k+n}=\ker f^k$ for some $n>0$.
$\ker f^k\subset \ker f^{k+n+1}$ is trivial. To prove the reverse inclusion, let $x\in\ker f^{k+n+1}$. So $f(x)\in\ker f^{k+n}$. By the inductive hypothesis, $f(x)\in\ker f^k$, i.e. $x\in\ker f^{k+1}=\ker f^k$ (initial step). The inductive step is proved.
