I know this statement is true if E is of finite dimension, but does it remain true in infinite dimension if we suppose that all the $\dim\ker f^n$ are finite ?
Here is my proof for finite dimension :
First, we have the two following identities : $$ \textrm{Im} f^{n+1}\subseteq \textrm{Im}f^{n}$$ $$ \dim \ker f^n + \textrm{Rank} f^n = \dim \ker f^{n+1} + \textrm{Rank} f^{n+1}$$ The third one implies $$ \dim\ker f^{n+1} - \dim\ker f^n = \mathrm{Rank} f^n - \mathrm{Rank}f^{n+1}$$ So it suffice to show that $\left(\mathrm{Rank} f^n - \mathrm{Rank}f^{n+1}\right)$ is decreasing.
To do that, let $d_n = \mathrm{Rank} f^n$, notice that, if we consider $g_n$ the restriction of $f$ to $\mathrm{Im} f^n$, then $\mathrm{Im}(f^{n+1}) =\mathrm{Im}( g)$ so that : $$ d_{n} = d_{n+1} + \dim\ker g_n$$ $$ \implies d_n - d_{n+1} = \dim\ker g_n$$ Because of the second identity, $\ker g_{n+1}\subseteq\ker g_n$ so that $\dim\ker g_{n+1} \le \dim\ker g_n$ and finally : $$ d_{n+1} - d_{n+2} \le d_{n} - d_{n+1}$$ Which proves the desired result.
This proof does not work in infinite dimension because we then have $\textrm{Im} f^n$ of infinite dimension for all $n$ because of rank-nullity theorem.
Thanks in advance for your help.
Note that for each $N\ge 1$, $\ker f^N$ is an invariant subspace of $f$, hence $$f_N=f\Big|_{\ker f^N}:\ker f^N\to \ker f^N$$ is well-defined. Since $\ker f^k =\ker f^k_N$ for $k\le N$ and $\ker f^N$ is assumed to be finite-dimensional, we can conclude the same as in the finite dimensional case.