A primitive and irreducible matrix is positive for some power $k.$ Prove it's positive for any power $k+i,$ where  $i=1,\dots.$ 
Let $P = [p_{ij}]_{1 \leqslant i,j \leqslant m} \geqslant 0$ a primitive and irreducible matrix. And $P^k > 0$ for some $k.$ Prove that $ P^{k+i} > 0, i =1,2, \dots.$ 

I have used a hint suggested by N.S below and wrote this proof what do you think?
By the irreducibility of $P \geqslant 0$ there exists a permutation matrix $M$ (an identity matrix whose rows and columns have been reordered) such that $MP$ can be written in the form $ MP = \begin{pmatrix} P_{11} & 0 \\ P_{21} & P_{22}  \end{pmatrix}$ where $P_{11}$ and $P_{22}$ are square matrices.  This implies $ P =  M^{-1}\begin{pmatrix} P_{11} & 0 \\ P_{21} & P_{22}  \end{pmatrix}.$ The matrix $P$  has at least a non-zero element (which is in fact positive) in each row, thus we use this fact in the proof below
We have $P^k >0.$ At $i=1$ we have $P^{k+1} = P P^k > 0.$ We assume at $i$ that we have $P^{k+i} >0.$ Thus at $i+1$ we have $ P^{k+i+1} = PP^{k+i} > 0.$ This is correct for all $i = 1,2, \dotsm$

Hint Prove first that any row of $P$ has a non-zero element. Then
$P^{k+i+1}=PP^{k+i}$
