Show that the function $v_n(x)\colon=\P^x(H_A > n)$ satisfies the recursion $v_{n+1}(x)=Pv_n(x)$ for all $x\in A^c$, $v_0=I_{A^c}$ and for all $n$ and all $x\in A$: $v_n(x)=0$.
\begin{eqnarray*}
v_n(x)
&=&\P^x(H_A > n)\\
&=&\E^x(\P^x(H_A\circ\Theta > n-1|\F_1);H_A\geq1)\\
&=&\E^x(\E^x(I_{[H_A > n-1}]\circ\Theta|\F_1);H_A\geq1)\\
&=&\E^x(\P^{X_1}(H_A > n-1);H_A\geq1)
=\E^x(v_{n-1}(X_1);H_A\geq1)~.
\end{eqnarray*}
For $x\in A^c$ we get: $v_n=Pv_{n-1}$ and for $x\in A$: $v_n(x)=0$