I know how to prove this using analytic techniques (just by using derivatives of $(1+z)^{\frac{1}{m}}$, and basic facts about power series), but I was wondering if there's any way to prove this using algebraic, or combinatorial techniques.
2026-03-27 09:48:09.1774604889
Algebraic or combinatorial proof that $(\sum_{n=0}^\infty {\frac{1}{m} \choose n} z^n )^m = 1+z$ as formal polynomials
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