Suppose $G$ is a $k-$regular graph and $A$ is its adjacency matrix, and let $[ \ 1 \ ]$ denote "all ones" square matrix of size $|V(G)|$. After some computations, I believe the following holds:
$$\lim_{n\to\infty} A^n/k^n=\frac{1}{|V(G)|}[ \ 1 \ ]$$
I sort of understand why, in terms of walks, but this intuition is not aiding me in the proof that this limit would require. Could someone point me in the right direction on how to prove this statement? Thanks.