There are b^n words of length n words over an alphabet with b letters. There is one empty word over any alphabet, including the empty alphabet. Also: every formula in combinatorics where 0^0 may occur (seriously, I'd love to see a single counter-example!). Your example says 0^0 = binomial(0,0) = number of ways to choose zero apples from a bag with zero apples = 1. All this of course over the integers. IIRC there was a discussion regarding 0^0 on math-fun a while ago. That discussion may have been discouraged at some point, though. Best regards, jj * Andres Valloud <ten@smallinteger.com> [Jun 13. 2020 15:28]:
I hear 0^0 should be defined as 1 for convenience, but perhaps there are proofs this has to be so (the *value* zero raised to the *value* 0, not the indeterminate limits of the form 0^0). What's your favorite proof that 0^0 = 1? Here's one.
0^0 = (1 - 1)^0 = \sum_{k=0}^0 \binom{0}{k} 1^{0-k} (-1)^k = \binom{0}{0} * 1 * 1 = \frac{0!}{0! \cdot 0!} * 1 = 1 * 1 = 1
Andres.
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