 Solution

Assume $r>0$, put $b=S^{\star }\in (0,K)$ and let ${\tau }_b=\inf \{u\ge 0:S_u\le b\}$. In the continuation region $S>b$, the time-independent Black-Scholes equation is a Cauchy-Euler equation. Substituting $V=S^m$ givesSet $\gamma =2r/{\sigma }^2>0$. The two powers are $S$ and $S^{-\gamma }$. The boundary condition $V(S)\to 0$ as $S\to \infty$ removes the growing solution; value matching at $b$ sets the coefficient. Hence the fixed-trigger value isFor $0\le S\le b$ the policy exercises immediately, giving $V_b(S)=(K-S{)}^n$. The quoted continuation expression alone does not apply below the exercise boundary. Equivalently, the discounted boundary-hitting representation gives ${\mathbb{E}}_S[e^{-r{\tau }_b};{\tau }_b<\infty ]=(S/b{)}^{-\gamma }$, which multiplies the exercise financial payoff.