Proof using the Markov property 
Consider a general chain with state space $S = {1,2}$ and write the transition probability as:   
Use the Markov property to show that $$P(X_{n+1} = 1) - \frac{b}{a+b} = (1-a-b) \bigl\{P(X_n = 1) - \frac{b}{a+b}\bigr\}$$
and then conclude $$P(X_n = 1) = \frac{b}{a+b} + (1-a-b)^n \bigl\{P(X_0 = 1) - \frac{b}{a+b}\bigr\}$$
  This shows that if $0 < a+b < 2$, then $P(X_n = 1)$ converges exponentially fast to its limiting value $b / (a+b)$.

I'm not really sure how to begin this proof. I've just begun to learn about the Markov property, and I'm not really sure how to use it in this way. Any push in the right direction (however small) is greatly appreciated.

Hint: Use the total probability formula:
$P(X_{n+1}=1)=P(X_{n+1}=1|X_n=1)P(X_n=1)+P(X_{n+1}=1|X_n=2)P(X_n=2)$
Then you take $P(X_{n+1}=1|X_n=1)=1-a$ and $P(X_{n+1}=1|X_n=2)=b$ from your transition matrix, and use $P(X_n=2)=1-P(X_n=1)$
... and you are going to be left with an equation in terms of $P(X_{n+1}=1)$, $P(X_n=1)$, $a$, and $b$.
