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What is the value of the expression: $\sum_{k=0}^{d}b^{k}\left[\begin{array}{c} d\\ k \end{array}\right]\prod_{i=0}^{k-1}\left(c-b^{i}\right) $ for $ k = 0 $?

In particular, what is then the value of: $\prod_{i=0}^{-1}\left(c-b^{i}\right)$?

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  • $\begingroup$ $\prod_{i=0}^{-1}\left(c-b^{i}\right)$ is the empty product and by convention it evaluates to $1$. For a more familiar example using the same notation, we have that $$ 5^n = \prod_{i = 1}^n 5 $$ What (intuitively should) happens when $n = 0$? This doesn't quite hold up as a strict formal proof, because of what happens for negative indices. But still, raising something to the zeroth power is the only place most people have met the concept of multiplying "no numbers" together. $\endgroup$
    – Arthur
    Oct 22, 2013 at 9:20

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Its same as multiplying no numbers together so the series $$\prod_{i=0}^{-1}\left(c-b^{i}\right)=1$$ in fact this works for all negative $n$ so that $$\prod_{i=0}^{-n}\left(c-b^{i}\right)=1$$ because her too we aren't multiplying any numbers!!

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