In a table tennis tournament (in which no game ends in a tie), each of the $n$ participants played exactly once against each of the others. It is known that for every $k>2$, there are no $k$ players $J_1, J_2, \ldots, J_k$ such that $J_1$ defeated $J_2, J_2$ defeated $J_3, J_3$ defeated $J_4, \ldots, J_{k-1}$ defeated $J_k, J_k$ defeated $J_1$.
Prove that there exists a player who defeated all the others, and there exists a player who lost to all the others.
Attempt:
Conjecture: For all $n \in \mathbb{N}_{\geq 2}$, the statement is valid.
Let's represent the problem as a directed graph, where $\vec{j_1 j_2}$ indicates that $j_1$ defeated $j_2$.
Initial cases, $n=2, n=3$. Assume it holds for $k \in \mathbb{\geq3}$.
Thus, consider that we have vertices $j_1, j_2,..., j_k, j_{k+1}$ such that $OUT(j_1) \leq OUT(j_2) \leq OUT(j_3) \leq .... \leq OUT(j_{k+1})$.
Removing $j_{k+1}$ and its edges, we have by assumption in the new graph: $$0 = OUT(j_1) \leq OUT(j_2) \leq OUT(j_3) \leq .... \leq OUT(j_{k})=k-1$$
Therefore, upon adding $j_{k+1}$ back, we have 2 cases.
Case 1) $OUT(j_{k+1})=k-1$
However, this implies that $j_{k+1}$ lost to $j_i \in {1, 2, \ldots, k}$.
If $i=1$
By hypothesis, this cycle cannot occur.
If $k>i>1$
By hypothesis, this cycle cannot occur.
If $i=k$, $OUT(j_k)=k> OUT(j_{k+1})$, which is absurd.
Therefore, only the second case remains, where $OUT(j_{k+1})=k$ and consequently $OUT(j_1)=0$.
I'm right?