$\ker T=\ker T^2 \implies\text{Im}\;T =\text{Im}\;T^2$ 
Assume $V$ is finite dimensional and let $T:V \to V$ be a linear operator. If $\ker T=\ker T^2$, then $\text{Im}\;T =\text{Im}\;T^2$

I am so confused about this proof. I am only able to know that if $\ker(T) = \ker(T^2)$ then $\dim(\text{Im}(T)) = \dim(\text{Im}(T^2))$, because of the dimension theorem. But how do we know $\text{Im}(T)$ and $\text{Im}(T^2)$ are exactly the same ? 
 A: Hint: There is a relationship between $\text{Im }T$ and $\text{Im }T^2$ which holds true even without the kernel condition. Think about what it means for a vector to be in each.
A: Only for give a solution that avoids dimension formula:
First of all, we have that $\operatorname{ker}T \cap \operatorname{im}T = \{ 0\}.$ Let $x \in V$ such that $x=Ty$ for some $y \in V$ and $Tx=0.$ Then $T^{2}y=Tx=0$ so $y \in \operatorname{ker}T^{2} = \operatorname{ker}T$ and $x = Ty = 0.$
Moreover, $\operatorname{ker}T=\operatorname{ker}T^{n}$ for every natural $n.$ Clearly $\operatorname{ker}T \subset \operatorname{ker}T^{n}.$ We prove it by induction. The result is true for $n=2$ by hypothesis. Suppose that it is also true for $n=k.$ We will show the result for $n=k+1.$ Let $x$ such that $T^{k+1}x=0,$ then $T^{k}(Tx)=0$ so $Tx \in \operatorname{ker}T^{k} = \operatorname{ker}T$ so $T^2(x)=T(Tx)=0$ and $x \in \operatorname{ker}T^2= \operatorname{ker}T.$
Now we prove the equality $\operatorname{im}T = \operatorname{im}T^{2}.$ Clearly we have the inclusion $\operatorname{im}T^2 \subset \operatorname{im}T.$ Now if $x \in \operatorname{im}T$ we have two possibilities: $x \in \operatorname{ker}T$ or $x \not\in \operatorname{ker}T.$ The first case will give us $x=0$ and clearly $x=0=T^2(0)$ so $x \in \operatorname{im}T^2.$ If $x \not\in \operatorname{ker}T$ then $x \not\in \operatorname{ker}T^{n}$ for every natural, hence $T^nx\ne 0 \ \forall n \in \mathbb{N}$ and $x\ne 0.$ Since the vector space $V$ is finite dimensional, say $\operatorname{dim}V=m$ then there is a linear combination of the vectors $\{x,Tx,\dots ,T^mx \}$ (these are $m+1$ vectors): $$a_0x + a_1 Tx + \dots + a_mT^mx = 0 \ : \exists i \text{ with } a_i \ne 0.$$ First suppose that $a_0 \ne 0,$ and we will have $$x = T\bigg(-\dfrac{a_1}{a_0}x - \dfrac{a_2}{a_0} Tx - \dots -\dfrac{a_m}{a_0}T^{m-1}x \bigg)$$ and since $x \in \operatorname{im}T$ and $T^kx \in \operatorname{im}T,$ being $\operatorname{im}T$ a linear subspace, we conclude that $x$ is the image of an element of $\operatorname{im}T$ and we conclude that $x \in \operatorname{im}T^2.$ If $a_0$ were $0,$ then $$T(a_1x + a_2Tx + \dots + a_mT^{m-1}x) = 0$$ but $a_1x + a_2Tx + \dots + a_mT^{m-1}x \in \operatorname{im}T$ so must be $a_1x + a_2Tx + \dots + a_mT^{m-1}x = 0$ (this sum is an element of the kernel and the image). Now you proceed as before. Observe that some $a_i \ne 0$ and by definition a linear combination is a finite sum so the process will finish.
A: First of all, observe that Im $T^2\subseteq$ Im $T$ as $T^2x=T(Tx)$ for all $x\in V$. By hypothesis, $\dim Ker(T)=\dim Ker(T^2).$ Now by rank-nullity theorem, we know 
$$ rank(T)+nullity(T)=\dim V\\
rank(T^2)+nullity(T^2)=\dim V. $$
Thus, $rank(T^2)=rank(T)$ i.e.,
$$\dim Im \text{ }(T)=\dim Im\text{ }(T^2).$$
Since Im $T^2\subseteq$ Im $T$ and their $\dim$ are equal implies that  Im $(T)= $Im $(T^2)$. 
