I recently encountered a proof of Nash's theorem that was so straightforward that it surprised me enough to want to share it.

Preamble

Let's assume we have a n-player game where each player can select from a finite number of deterministic strategies. This is the case in something like chess, or poker.

A player plays a pure strategy by selecting a single deterministic strategy from their available set. A player plays a mixed strategy if they select some probability distribution over all their available strategies and select a strategy to play at the start of the game by sampling from their distribution.

We need to define S, a set of all strategy profiles, which are assignment of players to mixed strategies. There is also a payoff to each player for playing the strategy profile $s_i$, which we call $u_i(s)$

Nash's theorem

Nash's theorem says that for each player there is some strategy profile $s^*$ such that for each i

$$u_i(s) \leq u$$

for all s. The profile $s^*$ is called a Nash equilibrium

and has been used as a 'solution' concept in games. I have some issues with calling it a solution, but it is an equilibrium in the sense that it does not profit an individual player to deviate from playing the strategy profile.

A sketch of the proof

First of all, we need a