# Rigorous solution of Stackelberg leader-follower game with N-players?

In the classical two-player Stackelberg game (sequential Cournot) we have a linear demand function $$P = 1 - Q$$ where $$Q = \sum_0^2 q_i$$, and we assume homogenous production cost $$c$$. By starting from the "bottom" of the game where player 2 knows player 1's chosen output and best-response reacts to it, and then moving "upwards" assuming player 1 anticipates this response, the game can be shown to have equilibrium solutions:

$$q_1^\star = \frac{1-c}{2}$$ $$q_2^\star = \frac{1-c}{4} .$$

We can follow the same procedure with 3, 4 or more players and discover the pattern: $$q_n^\star = \frac{1-c}{2^n} .$$

I have been trying to work out a rigorous mathematical proof, say by induction or other methods, to show that this has to be the case, but not managed to find any. Can anybody help please?

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• A sketch of proof was given by Andy Skrzypacz on Quora: qr.ae/TUtWqN – Herr K. Dec 16 '18 at 17:14

Because each new player faces the residual demand curve of the prior player, therefore he always ends up producing half the quantity of the prior player.

• OP asks for a "rigorous solution", not an intuition. – Herr K. Dec 16 '18 at 17:16