If the image of an operator is closed, is the image of the powers of the operator also closed?

Say $$T$$ is a bounded linear operator in a normed space that maps to itself (Banach or Hilbert space is fine). If the image $$\text{Im}(T)$$ is closed, then is it true that $$\text{Im}(T^n)$$ is closed? If not, what is a counterexample?

I know there are some theorems for compact operators that make use of this, and it is true for compact, but I'm not sure if this is true in general.

Let $$H$$ a separable Hilbert space with orthonormal basis $$\{e_n\}$$. Define $$T$$ as the linear operator induced by $$Te_n=\begin{cases} e_{n+1},&\ n\ \text{ even } \\[0.3cm] \frac1n\,e_n,&\ n\ \text{ odd} \end{cases}$$ Let $$M=\overline{\operatorname{span}}\{e_n,\ n\ \text{ odd}\}$$. Then $$\operatorname{Im} T=M$$ is closed (note that $$T(M)\subset M$$ and $$T(M^\perp)=M$$, and that $$M$$ is closed by definition). But $$T^2e_n=\begin{cases} \tfrac1{n+1}\,e_{n+1},&\ n\ \text{ even}\\[0.3cm] \tfrac1{n^2}\,e_n,&\ n\ \text{ odd} \end{cases}$$ The operator $$T^2$$ is compact with infinite-dimensional range, so it has non-closed range.
• My thinking was to have $T=K+N$, with $K$ compact and $N$ nilpotent with closed range. Dec 5, 2021 at 20:34