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Let $x=(x_1,\ldots,x_n)\in[-1,1]^n$, $k=(k_1,\ldots,k_n)$ and $\mathbb{N}^n_0=\mathbb{N}^n \cup\{0,\ldots,0\}$. Define $ x^k$ by $$x^k=\prod_{i=1}^n x_i^{k_i}$$

Show that $$\sum_{k\in\mathbb{N}^n_0}x^k=\prod_{i=1}^n (1-x_i)^{-1}$$

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up vote 1 down vote accepted

The factors $(1-x_i)^{-1}$ can be expanded as convergent geometric series $1+x_i+x_i^2+\cdots,$ since $x_i\in[-1,1].$ Thus, $\prod_{i=1}^n(1-x_i)^{-1}=(\sum_k x_1^k)\cdots(\sum_k x_n^k).$ Expanding we get as summands all possible products $x_1^{k_1}\cdots x_n^{k_n}$ with $(k_1,\ldots, k_n)\in\Bbb N^n,$ where $\Bbb N$ is nonnegative integers.

There is some justification to be made about why expanding is equal to the original product of series.

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