Prove if $im(T^k) \ne 0$ for some positive integer k for a nilpotent transformation T, then $im(T^{k+1}) \subsetneq im(T^k)$

Question

For a nilpotent transformation T, show that if $im(T^k) \ne {0}$ for some positive integer k, then $im(T^{k+1}) \subsetneq im(T^k)$

It can be inferred that $T^k$ is non-zero, since $im(0)$ = ${0}$ and $im(T^k)$ is non-zero. If $T^{k+1} = 0$, then shouldn't $im(T^{k+1})$ be contained in $im(T^k)$ since {0} is contained in $im(T^k)$? I'm not sure how to prove this besides that problem either. It was suggested to use proof by contradiction, but I'm not sure how. I know I'd start with the assumption that $im(T^k)$ is non-zero and $im(T^{k+1})$ is contained in $im(T^k)$, and then show that there is a contradiction with that, but I cannot find one. I keep feeling like the question itself is faulty.

Answer 1

sounds false to me.

$T: V \to V$ so $TV \subset V$

So

$$ T^{k+1} V = T^k(TV) \subset T^{k} V. $$

Answer 2

If $\mathcal{R}(T^{k})\ne \{0\}$ and $\mathcal{R}(T^{k+1})=\mathcal{R}(T^{k})$, then $T : \mathcal{R}(T^{k})\rightarrow\mathcal{R}(T^{k+1})=\mathcal{R}(T^{k})$ is surjective and, hence, invertible on $\mathcal{R}(T^{k})$, which cannot happen because the restriction of $T^{k}$ to this subspace $\mathcal{R}(T^{k})$ is nilpotent. (Note, I'm assuming finite-dimensional spaces to conclude invertibility.)