= Solution
Put $w(S)=|S|!(n-|S|-1)!/n!$ for $S\subseteq N\setminus\{i\}$. Under the bijection $S\mapsto R=(N\setminus\{i\})\setminus S$, the weights satisfy $w(S)=w(R)$ and $N\setminus S=R\cup\{i\}$. Consequently
$$
\sum_Sw(S)v(N\setminus S)=\sum_Rw(R)v(R\cup\{i\}).
$$
Subtracting $\sum_Sw(S)v(S)$ proves the alternative <Shapley value> expression
$$
\boxed{\phi_i(v)=\sum_{S\subseteq N\setminus\{i\}}w(S)\bigl(v(N\setminus S)-v(S)\bigr).}
$$
For the <dual coalitional game> $v'(S)=v(N)-v(N\setminus S)$, its <marginal contribution> at predecessor set $S$ is
$$
v'(S\cup\{i\})-v'(S)=v(N\setminus S)-v(N\setminus(S\cup\{i\}))=v(R\cup\{i\})-v(R).
$$
The same weight-preserving complement bijection therefore gives $\phi_i(v')=\phi_i(v)$, and both equal the displayed expression. \b[The <Shapley value> is invariant under coalitional duality.] This <Shapley self-duality> calculation complements within $N\setminus\{i\}$, not within $N$ without accounting for the distinguished player.
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