 Solution

For $S>b$, maximize the coefficient $H(b)=b^{\gamma }(K-b{)}^n$ of $S^{-\gamma }$. Its logarithmic derivative isIt changes sign once, at $b=\gamma K/(\gamma +n)$. The perpetual power put option therefore has candidate valueThe same boundary follows from smooth fit, $-\gamma (K-b{)}^n/b=-n(K-b{)}^{n-1}$. To establish optimality among all stopping times, rather than only fixed barriers, we verify the candidate. Let $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$ be the infinitesimal generator. The maximum property of $H$ gives $V_n\ge g_n$ everywhere, and $(\mathcal{L}-r)V_n=0$ for $S>b$. For $S\le b$, set $y=S/(K-S)\le \gamma /n$. ThenThus the discounted candidate is a nonnegative supermartingale. The smooth fit removes the derivative-jump local time of a semimartingale term in the Itô formula. The optional stopping theorem and localization bound every exercise policy's discounted expected value by $V_n$. Immediate exercise attains it below $b$; the first hitting time of $b$ attains it above $b$. The terminal remainder vanishes since $0\le e^{-ru}{V}_n(S_u)\le K^n{e}^{-ru}\to 0$. This proves the displayed value.

For $n=1$ it reduces to the ordinary perpetual put option. Larger $n$ lowers the optimal trigger, because the more strongly curved financial payoff makes further downward moves more valuable. If $r=0$, under the risk-neutral measure the stock tends to zero almost surely, so the supremum is $K^n$, approached as the exercise time tends to infinity and not attained at a finite time when $S>0$. If $r<0$, the values at deterministic exercise times grow without bound; the perpetual value is infinite. These cases do not admit the positive-rate formula above.