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I can show that the kernel of $T^k$ must be a subspace of the kernel of $T^{k+1}$. Because it is a subspace, we can show that if their dimensions are equal, then the kernels are equal.

To do this I am trying to use the Rank-Kernel Theorem, that is I know that

$\dim V$=$\dim Ker(T^k)+\dim Im (T^k)$ and the same for $T^{k+1}$. Thus in order for the kernels to be equal, there must exist a $k$ such that $\dim Im(T^k) =\dim Im(T^{k+1}).$ This is where I get stuck, it seems intuitive that for large enough $k$ this would be true, but I do not know how to prove that this is true.

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You have already made the following crucial observation: $$\ker(T) \subseteq \ker(T^2) \subseteq \ker(T^3)\subseteq \dots \subseteq V$$

Assume to the contrary that the claim you are trying to prove is false, then we actually have

$$\ker(T) \subsetneq \ker(T^2) \subsetneq \ker(T^3)\subsetneq \dots$$ Thus $\ker(T)$ has dimension at least $0$, $\ker(T^2)$ has dimension at least $1$, $\ker(T^3)$ has dimension at least $2$ and so on.

Let $n:= \dim V < \infty$. Then $\dim\ker(T^{n}) \geq n+1$, but since $\ker(T^n)$ is a subspace of $V$, we have $\dim(\ker T^n) \leq \dim(V) = n$, which is a contradiction.

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