= Solution
{c}
Let $D_n(a)$ denote the <constant term>, retaining the dimension as a subscript when it changes. We use the <Good recurrence for the Dyson constant term>. For algebraically independent variables, <Lagrange interpolation polynomial> applied to the constant polynomial one gives
$$
1=\sum_{i=1}^n\prod_{j\ne i}\frac{t-X_j}{X_i-X_j}.
$$
At $t=0$, this becomes the rational identity
$$
1=\sum_{i=1}^n\prod_{j\ne i}\left(1-\frac{X_i}{X_j}\right)^{-1}.
$$
When every $a_i>0$, multiplying by the given <Laurent polynomial> cancels one factor in row $i$ of the product and gives
$$
F(X;a)=\sum_{i=1}^n F(X;a-e_i),\qquad D_n(a)=\sum_{i=1}^n D_n(a-e_i).
$$
This is a polynomial identity after cancellation; no choice of an expansion of a rational function is involved.
The boundary case is essential. If $a_i=0$, the factors in row $i$ are absent. The only factors involving $X_i$ are then $(1-X_j/X_i)^{a_j}$ for $j\ne i$, and all their powers of $X_i$ are nonpositive. To obtain power zero in $X_i$, one must select the <constant term> one from each of them. Taking the <constant term> in $X_i$ therefore deletes that variable and exponent:
$$
D_n(a)=D_{n-1}(a_1,\ldots,\widehat{a_i},\ldots,a_n)\qquad(a_i=0).
$$
For $n=1$, the empty product is one, so $D_1(a_1)=1$. The all-zero exponent vector also gives one.
Now put $M_n(a)=m!/\prod_i a_i!$, where $m=\sum_i a_i$. This <multinomial coefficient> obeys the same deletion rule for a zero exponent. For positive exponents,
$$
\sum_i M_n(a-e_i)=\sum_i\frac{a_i}{m}M_n(a)=M_n(a).
$$
Induction on $n$, and within each dimension on $m$, consequently determines $D_n$ uniquely and identifies it with $M_n$. This proves the <Dyson constant-term identity>:
$$
\boxed{D(a)=\frac{m!}{a_1!\cdots a_n!}=\binom{m}{a_1,\ldots,a_n}.}
$$
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