Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ then $R(T) \cap N(T) = \{0\}$

We have that $T^2(V)\subseteq T(V)$, so $R(T)=R(T^2)$ implies that $T^2(V)=T(V)$.

Now, for the rank nullity theorem we know that $\dim V= R(T) + N(T)=R(T^2)+N(T^2)$, so $N(T)=N(T^2)$, but similarly to before, $\ker T \subseteq \ker T^2 $ and $\ker(T)=\ker(T^2)$.

Let $x \in \ker (T) \cap T(V)$. The fact that $x\in T(V)$ implies that there exists $x' \in V$ such that $T(x')=x$, then $T(T(x'))=T(x)=0$ (because $x \in \ker (T)$).

Then $x' \in \ker (T^2)=\ker (T)$, so $T(x')=0=x$.

We conclude that $\ker (T) \cap T(V)=\{ 0\}$.

You have $T^2(V)\subseteq T(V)$, and the dimension of these two subspaces is equal so $T(V) = T^2(V)$. Hence, $T$ is 1-1 on $T(V)$.