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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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    $\begingroup$ A sketch of proof was given by Andy Skrzypacz on Quora: qr.ae/TUtWqN $\endgroup$ – Herr K. Dec 16 '18 at 17:14
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Let $c = 0$ and $Q_k = \sum_{i = 1}^k{q_i}$. For $j = n$ the best response is given by \begin{align} b_n(Q_{n-1}) = \arg\max_{q_n}(1 - Q_{n-1} - q_n)q_n = \frac{1- Q_{n-1}}{2}. \end{align} For $j = n-1$ the best response is given by \begin{align} b_{n-1}(Q_{n-2}) =& \arg\max_{q_{n-1}}(1 - Q_{n-2} - q_{n-1} - b_n(Q_{n-1}))q_{n-1} = \frac{1- Q_{n-2}}{2}. \end{align} It can be shown that for any $k \in \{1,\ldots,n\}$ the best response is thus given by \begin{align} b_k(Q_{k-1}) = \arg\max_{q_k}\left(1 - Q_{k-1} - q_k - \sum_{\ell=k+1}^n{b_\ell}(Q_{\ell-1})\right)q_{n-1} = \frac{1- Q_{k-1}}{2}. \end{align} Note that $q_1^* = b_1(0) = \frac{1}{2}$. It then follows \begin{align} q_n^* = b_n(Q^*_{n-1}) =& \frac{1 - (q_1^* + q_2^* + q_3^* + \ldots + q_{n-1}^*)}{2}\\ =& \frac{1 - \left(q_1^* + \frac{1 - q_1^*}{2} + \frac{1 - q_1^* - q_2^*}{2} + \ldots + \frac{1 - (q_1^* + q_2^* + q_3^* + \ldots + q_{n-2}^*)}{2}\right)}{2}\\ =& \frac{1 - \left(q_1^* + \frac{1 - q_1^*}{2} + \frac{1 - q_1^* - \frac{1 - q_1^*}{2}}{2} + \ldots + \frac{1 - \left(q_1^* + \frac{1 - q_1^*}{2} + \frac{1 - q_1^* - q_2^*}{2} +\ldots + q_{n-2}^*\right)}{2}\right)}{2}\\ \vdots\\ =& \frac{1-q_1^*}{2^{n-1}}\\ =& \frac{1}{2^n} \end{align}

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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.

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

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