# Graph theory cycle proof

Let $k \geq 2$, let $G$ be a graph. Prove that if every vertex at least $k$, then $G$ contains a cycle of length atleast $k+1$.

I thought about proving this with a contradiction but I can't find a proof that proves that there exists no such cycle with length atleast $k+1$. I know there's a proof stating that if every vertex has degree at least $2$ then there exists some cycle, is this related to that proof in any way?

I assume the necessary condition that $G$ has finitely many vertices. Select an arbitrary vertex, $v_1$, and select any adjacent vertex, $v_2$. Then from the vertices connected to $v_2$, select any vertex not equal to $v_1$ and call it $v_3$. Continue in this fashion, each time selecting a vertex connected to $v_{n}$ not in the set $\{v_1,\ldots, v_{n-1}\}$. You can do this guaranteed at least until $n=k$ as each vertex has degree at least $k$. After this point the path has length $k-1$ as desired, so select another vertex connected to $v_k$ not in $\{v_2,\ldots, v_k\}$ and continuing not allowing for $v_n$ anything from $\{v_{n-k+1},\ldots, v_{n-1}\}$, and continue until you complete a cycle. The length is at least $k$ by construction.