Define $1_A$ to be the set of players whom player $A$ has beaten, $2_A$ to be the set of players not in $1_A$ but beaten by someone in $1_A$. We first show that an excellent player always exists. Suppose otherwise, and pick $A$ such that $|1_A\cup 2_A|$ is maximal. There exists $B\neq A, B\notin 1_A\cup 2_A$, by assumption. Since $B\notin 2_A$, $B$ must beat all of $1_A$. Also, $B\notin 1_A$, so $B$ beats $A$. Hence $(1_B\cup 2_B)\supseteq (1_A\cup 2_A\cup \{A\})$, contradicting the maximality. Now we suppose $A$ is the only excellent player. If $A$ does not beat everyone else, then $2_A$ is non-empty. Pick $B\in 2_A$ such that $B$ is excellent within $2_A$. Since $B\in 2_A$, $B\notin 1_A$ which implies that $B$ beats $A$, thus $1_A\subseteq 2_B$. We conclude that $B$ is (universally) excellent as well, a contradiction.