Prove that $T^2$ is the zero transformation if and only if $\mathrm{range}(T) \subseteq \ker(T)$.

$\newcommand{\range}{\mathrm{range}}$The full question:

$T^2$ means the composition, i.e. $T(T(x))$

Let $V$ be a vector space and let $T: V \rightarrow V$ be linear. Prove that $T^2$ is the zero transformation if and only if $\range(T) \subseteq \ker(T)$.

So far I know that for ($\implies$). Assume $T^2$ is the zero transformation. Then $\ker(T^2)=\{0 \in V: T^2(v)=0\}$. Note that this is $\forall v \in V$. We also know that the $\range(T)=\{0\in V: \exists v \in V: T^2(v)=0\}$.

Not sure where to go from here. Also for the other way, I don't know where to go.

Suppose that $\operatorname{ran}(T)\subseteq\ker(T)$. Then for any $x\in V$, $$T^2(x)=T\big(T(x)\big)\in T[\operatorname{ran}(T)]\subseteq T[\ker(T)]=\{0\}\;,$$ so $T^2(x)=0$.

Conversely, suppose that $T^2$ is the zero transformation, and let $x\in\operatorname{ran}(T)$. Then $x=T(y)$ for some $y\in V$, and $T(x)=T^2(y)=0$, so ... ?

This is very straightforward.

Suppose $T^2$ is the zero map. Then for any $v \in V$, $T^2(v) = T(T(v)) = 0$, which implies that $T(v) \in \ker T$ for any $v$. Hence $\{T(v): v \in V\} = T(V) \subset \ker T$. Now suppose, $T(V) \subset \ker T$. For any $v \in V$, $T^2(v) = T(T(v)) = 0$, so $T^2$ is the zero map.

$\newcommand{\range}{\mathrm{range}}$Hint: Assume by negation that $\range(T)\not\subset\ker(T)$ and take $v\in \range(T)$ s.t $v\not\in \ker(T)$.

What can you say about $T(v)$ ? can you find a vector $u$ s.t $T^{2}(u)\neq0$ ?

The other direction should be straightforward using the definitions (of composition, $\ker(T)$ etc')