Suppose the following situation: this is found at (Let $G$ be a graph of minimum degree $k>1$. Show that $G$ has a cycle of length at least $k+1$)
Let $P=v_0v_1 \dots v_l$ be a longest path in $G$. $v_0$ has to have additional neighbors by the degree constraint. All of the neighbors of $v_0$ have to be in $P$, otherwise $P$ could be extended. Therefore $v_0$ has at least $k$ neighbors in $P$. Let $j$ be the maximum index of a neighbor of $v_0$. By the previous statement we have that $j \ge k$. Thus we have the cycle $v_0v_1 \dots v_jv_0$, which has length at least $k+1$.
After many interpretation I did not understand what he meant by $j$ is the maximum index of a neighbor. I assumed it to be directed graph ? In addition I could not understand why $j\geq k$ and why has length at least k+1. I tried but I could not. ?