Let $V$ above $\mathbb{F}$ and let $T:V\to V$. Prove there is a $n\le \dim_\mathbb{F} V$ such that $V=\Im(T^n)\oplus \ker(T^n)$
Now I know that for all $k$: $\ker (T^{k}) \subseteq \ker(T^{k+1})$ and $\Im (T^{k+1}) \subseteq \Im (T^k)$. So if we look at $\ker T^0 \subseteq \ldots \subseteq \ker T^{n+1}$, there must be $i$ such that $\ker T^i = \ker T^{i+1}$.
Hence, there's no $v$ such that $T^i(v) \ne 0$ and $T^{i+1}(v) = 0$.
And this means, $\Im T^i \cap \ker T = \{0\}$
How should I proceed?
You are nearly there, continuing your idea there exists an $i$, such that $\text{ker} T^i = \text{ker} T^{i+1} = \text{ker} T^{i+2}=...=\text{ker} T^{i+k},\forall k \geq 0$. Therefore for $k=i$ holds: There does not exist $v$ such that $T^i v \neq 0$ but $T^{2i} v = 0$ and you get $\text{Image} T^i \cap \text{ker} T^i = \{0\}$. Now by the rank-nullity theorem, you conclude the desired claim.