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As part of a proof that $$(1-x)^n\cdot \left ( \frac{1}{1-x} \right )^n=1$$ in the context of generating functions it states that $$\sum_{i=0}^k(-1)^i\binom{n}{i}D(n,k-i))=0$$ where $D(n,k)$ is defined as $$D(n,k)=\binom{n-1+k}{k}=\binom{n-1+k}{n-1}\;.$$

I don't understand the above step, which is the last in the proof.

$$(1-x)^n=\sum_{i=0}^{\infty}(-1)^i\binom{n}{i}x^i$$

and

$$\left ( \frac{1}{1-x} \right )^n=\sum_{i=0}^{\infty}D(n,i)x^i$$

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    Could someone please explain to me why this requires a proof? Surely $(a)^{n} (b)^{n} \equiv (ab)^{n}$, so if $a=1-x$ and $b=\frac {1}{1-x}, ab=1$ for all x? Apologies if I just don't understand.2012-03-18
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    @Dan: I think the idea is just to verify that a direct proof is possible.2012-03-19
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    Satisfied by an answer below?2012-06-17

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