Gambler's ruin: verifying Markov property 
Gambler's ruin: the gambler starts with $\$i$, where $ 1<i<N$. He wins $\$1$ with probability $p$ and loses $\$1$ with probability $1-p$. When he reaches $0$ (ruin) or $N$ (win), he stops playing.

Durrett's book on Stochastic Processes states:

Let $X_n$ be the amount of money after $n$ plays. For any possible history of your wealth, $i_0, \ldots , i_{n-1},i,$ $$ P(X_{n+1}=i+1 | X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)=p $$
  since to increase your wealth by $1$ unit you have to win the next bet.

Although I see the point, I don't know how to prove the Markov property:
$$ P(X_{n+1}=j | X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)=P(X_{n+1}=j | X_n=i) $$
given the following hypothesis for the probability distribution of the sequence $\{X_n\}$:


*

*If $1<i<N$, then $P(X_{n+1}= i+1 | X_n=i)=p$ and $\ P(X_{n+1}=i-1 | X_n=i)=1-p$

*If $i=1$ or $i=N$, then $P(X_{n+1}= i | X_n=i)= 1$
How do I prove that the Markov property holds?
Edit: the states space is $\{0, \ldots, N \}$ and the probability distribution for $X_0$ is considered given 
 A: You can see that Markov property holds, using Durret's condition:

Let $X_n$ be the amount of money after $n$ plays. For any possible history of your wealth, $i_0, \ldots , i_{n-1},i,$ $$ P(X_{n+1}=i+1 | X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)=p $$
  since to increase your wealth by $1$ unit you have to win the next bet.

Indeed, writing:
$$P(X_{n+1}= i+i | X_n=i) = \frac{P(X_{n+1}= i+i , X_n=i)}{P(X_n=i)} =  \frac{\sum_{i_k, k \in \{1,\ldots,n-1\}}P(X_{n+1}=i+1 , X_n=i, X_{n-1}=i_{n-1}, \ldots, X_0=i_0)}{\sum_{i_k, k \in \{1,\ldots,n-1\}}P( X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)} = \frac{\sum_{i_k, k \in \{0,1,\ldots,n-1\}}p\cdot P( X_n=i, X_{n-1}=i_{n-1}, \ldots, X_0=i_0)}{\sum_{i_k, k \in \{1,\ldots,n-1\}}P( X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)} \\
=p\frac{\sum_{i_k, k \in \{1,\ldots,n-1\}} P( X_n=i, X_{n-1}=i_{n-1}, \ldots, X_0=i_0)}{\sum_{i_k, k \in \{1,\ldots,n-1\}}P( X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)} = p$$
where we considered the set of every possible outcomes in the $(n-1)$ bets in the sum and used that
$$  p = P(X_{n+1}=i+1 | X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0) = \frac{P(X_{n+1}=i+1 , X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)}{P( X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)}$$
implies
$$ p\cdot P( X_n=i, X_{n-1}=i_{n-1}, \ldots, X_0=i_0) \\= P(X_{n+1}=i+1 , X_n=i , X_{n-1}=i_{n-1}, \ldots, X_0=i_0)  $$
