A square matrix $A$ is such that $\mathrm{rank}(A^{k}) = \mathrm{rank}(A^{k+1})$ 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.
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.
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