= Solution
Apply $h$ to both sides of $\mathcal R(h^m)=\mathcal R(h^{m+1})$. Because $h(\mathcal R(h^j))=\mathcal R(h^{j+1})$, we obtain $\mathcal R(h^{m+1})=\mathcal R(h^{m+2})$. Repeating this argument proves equality at every later step. More formally, an <induction> on $k$ gives
$$
\boxed{\mathcal R(h^{m+k})=\mathcal R(h^m)\quad\text{for all }k\geq0.}
$$
The base case is $k=0$; the induction step follows from applying the <linear map> $h$ and then the assumed first equality. Thus one equality permanently stabilizes the descending sequence of <images of a linear map>.
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