If $P^r$ has all positive entries, then so does $P^n$ 
Let $P$ be the transition probability matrix of a Markov Chain. Argue
  that it for some positive integer r, $P^r$ has all positive entries,
  then so does $P^n$, for all integers $n\geq r$

I know that $$p_{ij}=P(X_{n+1}=j|X_n=i)$$ $$p(i,j)^r>0\space\forall i,j$$
If $r>0$ and $s>0$ such that $r+s=n$
$$p(i,j)^n=\sum_kp(i,k)^rp(k,j)^s>0\Rightarrow p(i,j)^n>0$$
Is that right? I'm bad when it comes to prove things.
 A: We want to see that
$$\forall\; (i,j):\;p(i,j)^r>0 \Rightarrow \forall(i,j)\;:\;p(i,j)^{r+1}>0$$
If  $\sum_k p(k,j) = 0$ then $p (l,j) = 0$ for every $l$ therefore $p(k,j)^r = \sum_{l}p(k,l)^{r-1}p(l,j) = 0$ which is a contradiction.
Therefore $ \exists\, k^*: p(k^*,j) > 0 $ 
and we conclude that $$ p_(i,j)^n \geq p(i,k^*)^rp(k^*,j)>0$$
Note: Use induction to get the general case.
A: Let S be the set of all the states of the Markov chain. Consider a matrix $A$ and $B$ such that $A$=$P^{r-1}$=$(a_{ij})_{i,j\in S}$ and $B$=$P^r$=$(b_{ij})_{i,j\in S}$ respectively, where $P$=$(p_{ij})_{i,j\in S}$ is the transition matrix of the Markov Chain. 
Then, 
$P^{r+1} = BP = \Biggl($$\sum_{k}  b_{ik}p_{kj} \Biggr)_{i,j\in S}$
As per the question, $(b_{ik})_{i,k\in S}>0$. So, for the summation to be equal to zero it must be that $p_{kj}=0$ $\forall$ $k\in S$. But since $P^r$= $\Biggl($$\sum_{k}  {a_{ik}}p_{kj} \Biggr)_{ij\in S}$ has positive entries, $p_{kj}\neq0$ $\forall$ $k\in S$ and hence, $P^{r+1}$ must have positive entries.
