Exercise :
Consider a random walk $\{ X_k \}_{k \in \mathbb N }$ on the integers with probability $p$ of going from $n$ to $n + 1$ and $1 - p$ of going from $n$ to $n - 1$. Let $T_{x-1}$ be a stopping time defined as $$T_{x-1} = \inf \{ k \ge 0 | X_k = x -1\}$$ Prove that $$\Pr [ T_0 < \infty | X_0 = x] = ( \Pr [T_0 < \infty | X_0 = 1] ) ^x$$
My attempt:
This walk can be modeled by the simple recurrence $$A(n,k + 1) = (1- p) A(n-1, k) + p A(n + 1, k) $$ where $n$ is the location and $k$ is the time index.
So I want to study $A(0, k)$ for the following two initial conditions
$A(1,0) = 1$
$A(x,0) = 1$
For the first case for example we clearly have that $A(0,1) = p$.
We can also model a second (scaled) chain where we walk $x -1$ steps each time either to the left or right so the recurrence is
$$B(n, k + 1) = (1-q) B( n - 1, k) + q B(n+1, k)$$ where $q = p^x$ so the second case transforms to the first one for $B(n,k)$ and we have that $B(0,1) = q = p^x$. But it is clear that $B(0,1) = A(0,k)$ for $A(x,0) = 1$ so the minimum probability of returning to 0 in finite time is the same for the two cases and equal to $p^x$. Its correctness relies on Strong Markov Property.
Am I following the correct way to solve it?
You have gone down an unnecessary alley when introducing the $x$-step chain.
Consider structuring your proof along these lines:
If $k>1$, then in order to eventually reach $k-1$ after starting at $k$, you must have traversed a $p$-random walk that is isomorphic with a $p$-walk starting at $1$ and ending at zero. Therefore, if $\tau(k)$ is the probability of terminating a walk starting at $k$, $$ \tau(k+1) = \tau(k) \tau(1) $$ From there, the closed form for $\tau(k)$ is easy, and the theorem you are proving follows trivially from that closed form.