Prove that markov chain is recurrent I have the following markov chain : 
$S=\{0,1,2,3\}$ 
$p_{i,0} = q$ (if we are in one of the states $0,1,2,3$ we can return to $0$ with probability $q$)
$p_{i,i+1} = 1-q , i\in\{0,1,2\}$ (if we are in state $i$ we can move to state i+1 with probability $1-q$)
$p_{3,3} = 1-q$ (state $3$ can move to itself with probability $1-q$)
$0<q<1$
Since this chain is finite and irreducible, either all states are recurrent or transient, so I only need to prove that one state is recurrent. I tried calculating $P^n$ . I found that the eigenvalues of $P$ are $1,0,0,0$, but I don't know how to proceed (or if there is a better way to prove that the state is recurrent?)
Any help will be welcomed.

Let $Y_t = 1 \text{ if }X_t \neq 0; Y_t = 0 \text{ if }X_t = 0$. This is a 2 states Markov chain; 0 is recurrent for $X$ iff it is recurrent for $Y$. 
For this Markov chain, the distribution of the time of return to 0 is a geometric law; it is almost always finite. Hence the chain is recurrent.
