If $\dim(Ker(f))$ is finite then so is $\dim(Ker(f^n))$ Since $f$ is a linear map and an endomorphism ($E \to E$)
and $n$ an integer $> 0$.
Hello. I would process by induction, but it is not that clear.
Thank you for you help. 
 A: Induction does the trick. Suppose $\ker f^n$ is finite-dimensional, choose a basis of $\ker f^n \cap \def\im{\mathop{\rm im}}\im f$, say $v_1, \ldots, v_\ell$ and a basis $u_{\ell+1}, \ldots, u_m$ of $\ker f$. Choose $u_i \in E$ such that $f(u_i) = v_i$ for $1 \le i \le \ell$. We will prove that $\{u_1,\ldots, u_m\}$ spans $\ker f^{n+1}$:
Now let $x \in \ker f^{n+1}$, that is $f(x) \in \ker f^n \cap \im f$. Therefore, there are $\lambda_1,\ldots, \lambda_\ell \in k$ such that $f(x) = \sum_{i=1}^\ell \lambda_i v_i$. Now $x - \sum_{i=1}^\ell \lambda_i u_i \in \ker f$, therefore we have $\lambda_{\ell+1}, \ldots, \lambda_m \in k$ such that 
$$ x - \sum_{i=1}^\ell \lambda_i u_i = \sum_{i=\ell+1}^m \lambda_i u_i \iff x = \sum_{i=1}^m \lambda_i u_i $$
Hence, $\ker f^{n+1}$ has a finite spanning set, therefore it is finite-dimensional.
A: By studying $f_{|f^{-1}(\ker f^{n-1})}\colon f^{-1}(\ker f^{n-1})\to \ker (f^{n-1}) $, we get from homomorphism theorem
$$(\ker(f^n))/(\ker(f))=(f^{-1}(\ker f^{n-1}))/(\ker f)\cong \ker f^{n-1}$$
So if $\dim (\ker f)<\infty$ we get by induction $\dim (ker (f^n))<\infty$.
