For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , show $V= \ker T^n \oplus $ Im $T^n$

I'm looking for the shortest and the clearest proof for this following theorem:

For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , I need to show $V= \ker T^n \oplus $ Im $T^n$.

Any hints? I don't know where to start from.

Thank you.

The main point is that ${\rm Ker}\ T^n \cap {\rm Im}\ T^n = \{0\}$. Show that if $T^k x \ne 0$ and $T^{k+1} x = 0$, then $x, Tx, \ldots T^k x$ are linearly independent, and therefore $k \le n-1$. No need here for Jordan Canonical Form - in particular this works over any field.

Let $E_1,...,E_i$ be the generalized Eigenspaces associated to $T.$ Show by examining the Jordan canonical form of $T$ that $T^n$ acts on $E_j$ as an automorphism if the eigenvalue associated to $E_j,$ denoted $\lambda(E_j),$ is non-zero and as the zero transformation otherwise. It follows $$\mathrm{Ker } T^n = \displaystyle\bigoplus_{\lambda(E_j) = 0} E_j$$ and $$\mathrm{Im} T^n = \displaystyle\bigoplus_{\lambda(E_j) \neq 0} E_j$$ and $$V = Ker T^n \oplus Im {T^n}.$$

If $S$ is an endomorphism of a finite dimensional vector space $V$, then $$V=\mathrm{Ker}\ S\ \oplus\ \mathrm{Im}\ S\Leftrightarrow\mathrm{Ker}\ S\ \cap\ \mathrm{Im}\ S=0\Leftrightarrow\mathrm{Ker}\ S^2=\mathrm{Ker}\ S.$$

Put $S:=T^n$.

I find Robert Israel's argument wonderful, but it seems to me that jspecter's proof works over any field.

Indeed, let $T$ be an endomorphism of an $n$-dimensional vector space $V$ over a field $K$. Let $f\in K[X]$ be the minimal of $T$. Write $f=X^rg$ with $g(0)\not=0$. By the Chinese Remainder Theorem, $K[T]=K[X]/(f)$ is the product of $K[X]/(X^r)$ and $K[X]/(g)$, yielding a decomposition $V=V_X\oplus V_g$, where $T$ is nilpotent on $V_X$ and invertible on $V_g$.