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 $V_g$.