= Solution
Because $A$ is nondecreasing and $A_{n+1}$ is $\mathcal F_n$-measurable,
$$
\{\tau^*\leq n\}=\{A_{n+1}>0\}\in\mathcal F_n.
$$
The convention $A_{N+1}=\infty$ ensures $\tau^*\leq N$, so $\tau^*$ is a bounded stopping time.
By minimality, $A_{\tau^*}=0$ and $A_{\tau^*+1}-A_{\tau^*}>0$. Part (d) then gives
$$
V_{\tau^*}=Z_{\tau^*}.
$$
Moreover $M_{\tau^*}=V_{\tau^*}+A_{\tau^*}=Z_{\tau^*}$. Optional sampling for the martingale $M$ now yields
$$
\boxed{
\mathbb E[Z_{\tau^*}]
=\mathbb E[M_{\tau^*}]
=M_0=V_0.}
$$
Together with part (a), this proves that $\tau^*$ is optimal and $V_0$ is the optimal stopping value.
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