Prove that if $D$ is a digraph such that $od(v) \geq k \geq 1$ for every $v \in V(D)$ then $D$ contain a cycle of length at least $k+1$
I tried to prove this by induction. So here is what I got so far
Base: $od(v)=k=1$, then I got a digraph with a bunch of symmetric arcs, which contain a lot of cycle length 2. So the base case is good.
Inductive: Suppose that the statement is true for some $k$, meaning $od(v) \geq k \geq 1$ for every $v \in V(D)$ then $D$ contain a cycle of length at least $k+1$. And I want to show $od(v) \geq k+1 \geq 2$ for every $v \in V(D)$ then $D$ contain a cycle of length at least $k+2$.
I add one vertex called $w$, since every vertex must have out degree at least $1$, connect $w \to v$, but it doesn't say that the in degree must be at least 1, so I don't know if this new vertex will increase the length of the cyle.
We consider the longest path $v_0 \to v_1 \to \cdots \to v_n$ where the $v_i \neq v_j$ for $i \neq j$. If there is a vertex $u \notin \{v_0, \cdots, v_n \}$, then $v_0 \to \cdots \to v_n \to u$ becomes a longer path. Hence for all $v_n \to u$, there exists an $i$ such that $u = v_i$.
Now because $od(v_n) \ge k$, there should exist at least $k$ vertices $v_{a_1}, \cdots, v_{a_k}$ such that $v_n \to v_{a_i}$ for all $i = 1, \cdots, k$ and $a_1 < \cdots < a_k < n$. Since $a_1 \le n-k$, we now have a cycle $v_{a_1} \to v_{a_1+1} \to \cdots \to v_n \to v_{a_1}$ which is of length at least $k+1$.