Binomial Theorem/Hurwitz's Generalisation

Theorem

 * $\displaystyle \left({x + y}\right)^n = \sum x \left({x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} y \left({y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}\right)^{n - \epsilon_1 - \cdots - \epsilon_n}$

where the summation ranges over all $2^n$ choices of $\epsilon_1, \ldots, \epsilon_n = 0$ or $1$ independently.

Proof
Follows from this formula:
 * $(1): \quad \displaystyle \sum x \left({x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}\right)^{\epsilon_1 + \cdots + \epsilon_n - 1} y \left({y + \left({1 - \epsilon_1}\right) z_1 - \cdots + \left({1 - \epsilon_n}\right) z_n}\right)^{n - \epsilon_1 - \cdots - \epsilon_n} = \left({x + y}\right) \left({x + y + z_1 + \cdots + z_n}\right)^{n - 1}$