a tournament is playing between n persons. Everybody plays with everybody one time. There is no draw here. A number $k$ is called $n$ good if there is any tournament such that in that tournament they have any player in the tournament that has lost all of $k$'s.
prove that
1. $n$ is greater than or equal to $2^{k+1}-1$
2.Find all $n$ such that $2$ is a n-good
1. **Prove that \( n \geq 2^{k+1} - 1 \)**

   We will use induction to prove this statement.

   **Base Case: \( k = 1 \)**

   For \( k = 1 \), we need to show that \( n \geq 2^{1+1} - 1 = 3 \). In a tournament with 3 players, it is possible to have a player who loses to both of the other players. Thus, the base case holds.

   **Inductive Step:**

   Assume the statement holds for \( k = m-1 \), i.e., \( n \geq 2^m - 1 \). We need to show that it holds for \( k = m \), i.e., \( n \geq 2^{m+1} - 1 \).

   Suppose, for the sake of contradiction, that there exists a tournament with \( n \leq 2^{m+1} - 2 \). In this tournament, there must be a player (call her Ruba) who has won at most \( \left\lfloor \frac{1}{2}(n-1) \right\rfloor = 2^m - 2 \) matches.

   There are two cases to consider:
   
   **Case 1: Ruba has won at least \( m-1 \) matches.**

   Let \( L \) be the set of players whom Ruba has defeated. By definition, \( |L| \) is at least \( m-1 \) and at most \( 2^m - 2 \). For any group of \( m-1 \) players in \( L \), there exists another player (call him Munna) in the tournament who has lost to each of these \( m-1 \) players and to Ruba. Munna must be a member of \( L \). Therefore, \( L \) is \( (m-1) \)-good, and by the inductive hypothesis, \( |L| \geq 2^m - 1 \), which is a contradiction.

   **Case 2: Ruba has won fewer than \( m-1 \) matches.**

   Choose \( m-1 - |L| \) players who are neither Ruba nor members of \( L \) and add them to \( L \) to form a set \( L' \) of \( m-1 \) players. By hypothesis, there is a player (say, Topu) who has lost to all members of \( L' \) and to Ruba. Topu must be a member of \( L \), but \( L \subset L' \), which is a contradiction since Topu cannot lose to himself.

   Therefore, \( n \geq 2^{m+1} - 1 \) must hold, completing the inductive step.

   \(\blacksquare\)

2. **Find all \( n \) such that \( 2 \) is \( n \)-good.**

   From part (a), we know that \( n \geq 2^{2+1} - 1 = 7 \) is necessary. We need to show that \( n = 7 \) is sufficient.

   Consider the case \( n = 7 \). Label the 7 players as \( p_0, p_1, \ldots, p_6 \). Define the tournament results as follows:
   \[
   p_0 \to p_1 \to p_2 \to p_3 \to p_4 \to p_5 \to p_6 \to p_0
   \]
   \[
   p_0 \to p_2 \to p_4 \to p_6 \to p_1 \to p_3 \to p_5 \to p_0
   \]
   \[
   p_0 \to p_4 \to p_1 \to p_5 \to p_2 \to p_6 \to p_3 \to p_0
   \]

   This tournament has the desired property that for every two players, there is a third player who has lost to both. This is motivated by the fact that the set \(\{1, 2, 4\}\) modulo 7 covers all residues.

   For \( n > 7 \), add \( n-7 \) players \( q_7, q_8, \ldots, q_{n-1} \) such that each game between a \( q \) and a \( p \) is won by the \( q \). The games between the \( q \)s can be assigned arbitrarily. This construction satisfies the problem statement.

   \(\blacksquare\)

The final answer is \( \boxed{ n \geq 7 } \).