Directed Cycles of strict Digraphs with minimum outdegree k (What is wrong with my solution?) The Question is:
Deduce, by induction on $n$, that every strict digraph $D$ with minimum outdegree $k$, where $k \geq 1$, contains a directed cycle of length at most $\frac{2n}{k}$.
My solution so far is :
The base case is for $n=3$, for which $k=1$ ($n=2$ has $k=0$). This oriented triangle has a directed cycle of at most length $3$, which is less than $\frac{2*3}{1} = 6$. Assume this holds up until $n-1$. Now consider a strict digraph on $n$ vertices with minimum outdegree $k$. Remove $x$ with minimum outdegree $k$ to get $D'$ with minimum outdegree $k$ **, and thus by the inductive hypothesis we get that there exist directed cycles of length no greater than $\frac{2n-2}{k}$, which holds for all $k$. Reinsert $x$, and now we have that every directed cycle in $D'$ exists in $D$ with the possible addition of traversing over $x$. Thus we have paths of length no greater than $\frac{2n}{k} + (1-\frac{2}{k})$.
** This uses the following theorem: Let $D$ be an oriented graph with minimum outdegree $k$, where $k \geq 1$. Let $D'$ be the digraph obtained from $D$ by deleting $N^{-}(x) \cup \{x\}$ and adding an arc $(u, v)$ from each vertex $u$ of the set $N^{--}(x)$ (where $x$ has minimum outdegree and indegree of at least $k$***) of in-neighbours of $N^{-}(x)$ to each vertex $v$ of $N^{+}(x)$, if there was no such arc in $D$. Show that $D'$ is a strict digraph with minimum outdegree $k$.
*** This uses the following theorem: Let $D$ be an oriented graph with minimum outdegree $k$, where $k \geq 1$. Show that $D$ has a vertex $x$ whose indegree and outdegree are both at \indent least $k$.
This is where I am stuck. The conclusion holds for $k=1$ and $k=2$, but i can't make the leap for larger values of $k$. What am I missing 
 A: Well, already given in the comments was why the OP's approach doesn't yet give the desired result. OP, I have no idea how you could modify your approach to solve the problem (sorry). This in all fairness seemed like a challenging exercise, and the below is the only way I know how to solve this. 
We assume the following: 

IH: Let $G'$ be a digraph on $n'<n$ vertices and minimum outdegree $k'$ for some positive integer $k'$. Then $G'$ has a directed cycle of length no more than $2n'/k'$.

Now pick an arbitrary vertex $v \in G$ and let $N^1(v)$ denote the set of vertices $w_1$ such that $vw_1 \in G$; $N^2(v)$ denote the set of vertices $w_2 \in G \setminus N^1(v)$ such that there exists a $w_1 \in N^1(v)$ where $w_1w_2 \in G$. And so on; for general $j$ let $N^{j+1}(v)$  denote the set of vertices $w_{j+1}$ in $G \setminus (N^1(v) \cup N^2(v) \cup \ldots N^j(v))$ such that there exists a $w_j \in N^j(v)$ where $w_jw_{j+1} \in $G$.

Claim 1: Now let $j_0$ be the smallest integer such that either $|N^{j_0+1}(v)| \le k/2$, OR $|N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)| > n/2$. (i) Then $j_0 \le n/k$. (ii) Furthermore, lest $G$ has a cycle of length $2n/k$ or less, the set $N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)$ must have more than $n/2$ vertices.

To see (i) note that $|N^1(v) \cup N^2(v) \cup \ldots N^{j+1}(v)| > k/2$ $+$ $|N^1(v) \cup N^2(v) \cup \ldots N^{j}(v)|$ for each $j<j_0$
so $|N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)| > j_0(k/2)$. However, $(n/k)(k/2) > n/2$ so $j_0$ can be no larger than $n/k$. 
We show (ii). Indeed, note that if $|N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)| \le n/2$, then every vertex in $G[N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)]$ has outdegree at least $k/2$. So if $|N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)|$ has $n/2$ vertices or fewer, then IH implies a cycle of length no more than $\frac{2(n/2)}{k/2} = 2n/k$ and we are done.
Letting $j_0$ be as in Claim 1, and letting $w$ be any vertex in $N^1(v) \cup N^2(v) \cup \ldots N^{j_0}(v)$, we say that $v$ covers $w$. So every vertex $v$ covers more than half of the vertices in $G$. So there must be another vertex $v'$ that is covered by more than half of the vertices in $G$. So there must in turn be a vertex $w'$ such that $w'$ is covered by $v'$ and $v'$ is covered by $w'$. [Indeed, let $S'$ be the set of vertices of $G$ covered by $v'$ and let $T'$ be the set of vertices of $G$ that cover $v'$. Then as both $S'$ and $T'$ have more than $n/2$ vertices and $G$ has only $n$ vertices, it follows that $S' \cap T'$ must be nonempty. Any $w' \in S'\cap T'$ will do.] But then this implies a directed path of length $n/k$ or less from $v'$ to $w'$ and in turn a dipath of length $n/k$ or less from $w'$ back to $v'$. This gives a directed cycle of length $2n/k$ or less in $G$--and by the fact that $G$ is strict, this is a directed cycle with 3 or more vertices.
