$f$ nilpotent and $\dim \ker f^k = \dim \ker f^{k+1}$ then $f^k=0$ Let $f:\mathbb R^n\rightarrow \mathbb R^n$ a linear nilpotent operator with $\dim \ker f^k = \dim \ker f^{k+1}$ for $k\in \mathbb N$. How can I prove that $f^k=0$?
I tried to show that because $f$ is nilpotent, there exists  $i \in \mathbb N$ with $f^i=0$. and we know that as we go further to $i$ the kernel of $f$ gets smaller and smaller. but It seems that something is missing here, I think that might be a much easier solution.

Notice that $\ker f^k\subset \ker f^{k+1}$. So, if $\dim\ker f^k=\dim\ker f^{k+1}$, than $\ker f^k= \ker f^{k+1}$.
Finally, for the sake of contradiction, assume there exists some $v$ such that $f^k(v)\neq0$
$$
f^k(v)\neq0\rightarrow f^{k+1}(v)\neq0\rightarrow
 f^k(f(v))\neq0\rightarrow
 f^{k+1}(f(v))\neq0\rightarrow f^{k+2}(v)\neq0\dotsb
$$
Hence, for every $n \in \mathbb{N}$, $f^n(v)\neq0$.
