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Let $D(a)$ denote the stated constant term and suppose every $a_i>0$. The rational identitycan be verified after clearing denominators, or by Lagrange interpolation. Multiplying it by the Dyson product and taking constant terms gives the recursionThe multinomial coefficientobeys the same recursion by the multinomial form of Pascal's identity.

If $a_k=0$, taking the constant term in $X_k$ forces the zero term from every factor involving $X_k$ and reduces the expression to the $(n-1)$-variable Dyson product with $a_k$ omitted. The same boundary reduction holds for $M(a)$. Finally $D(0,\dots ,0)=M(0,\dots ,0)=1$. Induction on $n$ and on $a_1+\cdots +a_n$ therefore proves the Dyson constant-term identity

Identify the additive group $G$ with the finite field ${\mathbb{F}}_q$, where $q=p^{\gamma }$. The sets $A$ and $B$ each have size $k<p$, and the field has odd characteristic. The Snevily matching theorem for an elementary abelian group states precisely that for two $k$-element subsets of the additive group of such a field, there is a bijection $\pi :A\to B$ for which the sums $a+\pi (a)$ are pairwise distinct.

For context, its polynomial proof antisymmetrizes the Vandermonde polynomialover all orderings of $B$. The coefficient furnished by the Dyson constant-term identity is a nonzero multiple of $k!$; it cannot vanish because $k<p$. Therefore at least one ordering $(b_1,\dots ,b_k)$ makes the Vandermonde product nonzero, which says exactly that