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Write $u={\sum }_{\lambda }{a}_{\lambda }e(\lambda )$ with finite support. The root reflection is an involution, and $s_{\alpha }u=-u$ implies $a_{s_{\alpha }\lambda }=-a_{\lambda }$. Any fixed weight has coefficient zero. For each nonfixed orbit choose its representative $\lambda$ withIntegrality of $m$ follows because $\lambda$ belongs to the weight lattice; positivity is ensured by our choice of orbit representative. Pairing the orbit coefficients givesFor each pair, elementary finite telescoping showsThere are finitely many chosen orbits, and each inner sum is finite. This proves the root-binomial divisibility of reflection anti-invariants:

For precision, the geometric-series expression is interpreted coefficientwise. Its coefficient at a weight $\nu$ is ${\sum }_{n\ge 0}{a}_{\nu -n\alpha }$, which is a finite sum because $u$ has finite support. On each coset of $\mathbb{Z}\alpha$, reflection pairs the coefficients with opposite signs, so their total is zero. These cumulative sums vanish both before the smallest occupied weight and after the largest occupied weight. Only finitely many cosets are occupied, hence the product of the formal series with $u$ has finite support and equals the boxed quotient.

The individual terms $e(n\alpha )u$ need not vanish for large $n$. For example, $u=e(\alpha )-e(-\alpha )$ is anti-invariant, and its quotient is $-1-e(-\alpha )$. Thus “finite sum” means a finite result after coefficientwise cancellation, not an actually truncated geometric series. No analytic convergence assumption is involved.