A square matrix $A$ is such that $\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+1})$. Then $\mathrm{rank}(A^{k})=\mathrm{rank}(A^{k+i})$ for all $i$.
I think we prove this by induction on $i$.
Let $i = 2$. We show that rank $(A^{k})$ = rank $(A^{k+2})$.
Now, we know that rank $(A^{k+2})$ = rank $(A^{k+1}A)$ $\le$ min (rank$(A^{k+1})$, rank$(A)$) $\le$ rank $(A^{k+1})$ = rank $(A^{k})$.
However, I cannot proceed further.
On
Think geometrically. We have $A^{k+1}=A^kA$, so the image of $A^{k+1}$ is contained in the image of $A^k$. If the rank of the two are equal, that means the image is the same. Since $A^{k+1}=AA^k$, this means that $A$ is a bijection on said image.
On
This can be nicely shown by thinking of $A$ as a linear transformation $f \colon K^n \to K^n$ and looking at the decreasing chain $\operatorname{im} f^p \supseteq \operatorname{im} f^{p+1}$ as follows:
If $A \in \operatorname{Mat}_n(K)$ then let $f \colon K^n \to K^n$ with $f(x) = Ax$ be the associated endomorphism of $K^n$, and let $R_p := \operatorname{im} f^p$ for all $p \geq 0$. Then $$ \operatorname{rank} A^p = \dim \operatorname{im} f^p = \dim R_p \quad \text{for all $p \geq 0$}, $$ as well as $$ f(R_p) = f(\operatorname{im} f^p) = \operatorname{im} f^{p+1} = R_{p+1} \quad \text{for all $p \geq 0$}. $$
Because we have a decreasing chain $$ K^n = R_0 \supseteq R_1 \supseteq R_2 \supseteq R_3 \supseteq \dotsb $$ it follows from $\operatorname{rank} A^{k+1} = \operatorname{rank} A^k$ that $\dim R_{k+1} = \dim R_k$ and thus $R_{k+1} = R_k$.
It then further follows that $$ R_{k+2} = f(R_{k+1}) = f(R_k) = R_{k+1}, $$ and we find inductively that $R_{k+j} = R_{k+j+1}$ for all $j \geq 0$. So we have $$ R_k = R_{k+1} = R_{k+2} = R_{k+3} = \dotsb $$ and thus $R_{k+i} = R_k$ for all $i \geq 0$.
so you know now that $rank (A^k)=rank (A^{k+1})=rank (A^{k+2})$. you just can take $k'=k+1$ and do the same work for $A^{k'}$. You obtain $$rank (A^k)=rank (A^{k+1})=rank (A^{k'})=rank (A^{k'+1})=rank (A^{k'+2})= rank (A^{(k+1)+2}) = rank (A^{k+3}) $$ and so on