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
$$ V = \ker(T^{n-1}) \oplus \operatorname{im}(T^{n-1}). $$
Best Answer
He started with an arbitrary vector $v + \operatorname{null} T\in \operatorname{null} S$ and showed that the vector must be the zero vector of $V/\operatorname{null} T$, i.e., $0 + \operatorname{null} T$. Thus, the only possible element of $\operatorname{null} S$ is $0 + \operatorname{null} T$. Since $0 + \operatorname{null} T$ is already in $\operatorname{null} S$, you deduce that $\operatorname{null} S = \{0 + \operatorname{null} T\}$ (here he simply writes $\operatorname{null} S = 0$).