Show that $V=\operatorname{null} T^{n-1}\oplus \operatorname{range} T^{n-1}$ This question comes from A. Sheldon's Linear Algebra Done Right, 3rd Edition, Exercises 8.A.

Suppose that $T\in\mathcal{L}(V)$ is not nilpotent. Let $n=\dim V$. Show that $V=\operatorname{null} T^{n-1}\oplus \operatorname{range} T^{n-1}$.

After pondering over and over this question for a long time, I've tried expanding basis, constructing quotient space, and many others, but I cannot unravel this stuff. Unfortunately, my university hasn't adopted this excellent textbook for linear algebra course, and I have to learn it on my own. I've met many hard exercises in this book but this one seems to defy all my attempts.
Any help or hint is welcome, best in the manner of "Axler's way", thank you so much.
 A: Let's prove something slightly more general. 
Assume that $V$ is a finite dimensional vector space and $T \in \mathcal{L}(V)$. We have the nested chain of ascending subspaces
$$ \{ 0 \} = \ker(T^0) \subseteq \ker(T) \subseteq \ker(T^2) \subseteq \dots $$
and since $V$ is finite dimensional, this chain must stabilize after finitely many steps. In fact, it must stabilize after at most $n$ steps (where $n = \dim V$). Let $N \in \mathbb{N}_0$ be the minimal number such that $\ker(T^N) = \ker(T^{N+1})$ (so $N \leq n$). Then the chain of subspaces looks like
$$ \{ 0 \} = \ker(T^0) \subsetneq \ker(T^1) \subsetneq \dots \subseteq \ker(T^N) = \ker(T^{N+1}) = \ker(T^{N+2}) = \dots $$
We also have the nested chain of descending subspaces
$$ V = \operatorname{im}(T^0) \supseteq \operatorname{im}(T) \supseteq \operatorname{im}(T^2) \supseteq \dots $$
Since $\dim \ker(T^i) + \dim \operatorname{im}(T^i) = n$, this chain also stabilizes precisely after $N$ steps so we also have
$$ V = \operatorname{im}(T^0) \supsetneq \operatorname{im}(T) \supsetneq \dots \supsetneq \operatorname{im}(T^N) = \operatorname{im}(T^{N+1}) = \operatorname{im}(T^{N+2}) = \dots $$
Let's show that $V = \ker(T^N) \oplus \operatorname{im}(T^N)$. Since $\dim \ker(T^N) + \dim \operatorname{im}(T^N) = n$, it is enough to show that $\ker(T^N) \cap \operatorname{im}(T^N) = \{ 0 \}$. Let $T^N(v) \in \ker(T^N) \cap \operatorname{im}(T^N)$ so $T^{N}(T^N(v)) = T^{2N}(v) = 0$. Since $2N \geq N$, we get $v \in \ker(T^{2N}) = \ker(T^N)$ so $T^N(v) = 0$. Note that since the chains stabilize after $N$ steps, we also have
$$ V = \ker(T^i) \oplus \operatorname{im}(T^i) $$
for all $i \geq N$.

Now let's get back to your problem. If the chains above stabilize at $N = n$ then 
$$ n \geq \dim \ker(T^n) > \dim \ker(T^{n-1}) > \dots > \dim \ker(T^0) = 0$$
which implies that in fact $\dim \ker(T^n) = n$ so $T^n = 0$ and $T$ is nilpotent. Hence, if $T$ is not nilpotent, $N \leq n - 1$ and so we have
$$ V = \ker(T^{n-1}) \oplus \operatorname{im}(T^{n-1}). $$
A: Hint: Here is a roadmap:


*

*$V=\operatorname{null} T^{n}\oplus \operatorname{range} T^{n}$, for all $T$. 

*If $T$ is nilpotent, then $T^n=0$.
The main tool is this:


*

*$\operatorname{null}T^k\subseteq\operatorname{null}T^{k+1}$ and 
$\operatorname{range}T^{k+1}\subseteq\operatorname{range}T^k$, for all $k$.

*Both sequences stabilize at the same $k$.

