In 1970, Herbert Vaughan [10] argued for the explicit recognition of evaluating 0^{0} = 1. He aimed to show "that there is a good deal of motivation for defining '0^{0}' to be a numeral for 1." He provided three examples.
Example 1. Vaughan gave the infinite geometric progression
\[\sum_{n=1}^{\infty} x^{n1} = \frac{1}{1x} \mbox{ for }  x  < 1.\]

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If \(x = 0,\) then \(\vert x\vert = \vert 0\vert < 1,\) which leads to
\[\sum_{n=1}^{\infty} 0^{n1} = \frac{1}{10} = 1.\]

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The infinite sum can be expanded as 0^{0} + 0^{1} + 0^{2} + … = 1. As stated by Vaughan, if 0^{0} is not defined, this summation is senseless. Further, if 0^{0} ≠ 1, then the summation is false.
Example 2. This example arises from the infinite summation for e^{x}, which can be written as
\[\sum_{n=1}^{\infty} \frac{x^{n1}}{(n1)!} = e^x \mbox{, for all } x.\]

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Everyone agrees that 0! = 1, so in the case where x = 0, the sum becomes
\[\sum_{n=1}^{\infty} \frac{0^{n1}}{(n1)!} = e^0 = 1.\]

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The sum can be expanded as
\[\frac{0^0}{0!} + \frac{0^1}{1!} + \frac{0^2}{2!} + \cdots = \frac{0^0}{1} + 0 + 0 + \cdots = 0^0.\]

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The righthandside of the summation is e^{0} = 1, so 0^{0} = 1.
Example 3. A third example given by Vaughan involves the cardinal number of a set of mappings. In set theory, exponentiation of a cardinal number is defined as follows:
a^{b} is the cardinal number of the set of mappings of a set with b members into a set with a members.
For instance, 2^{3} = 8 because there are eight ways to map the set { x, y, z } into the set { a, b }. In order to calculate 0^{0}, determine the number of mappings of the empty set into itself. There is precisely one such mapping, which is itself the set of the empty set. "So, as far as cardinal numbers are concerned," wrote Vaughan, "0^{0} = 1."
When might a mathematician want 0^{0} to be something that is not indeterminate? If, for example, we are discussing the function f(x, y) = x^{y}, the origin is a discontinuity of the function. No matter what value may be assigned to 0^{0}, the function x^{y} can never be continuous at x = y = 0. Why not? The limit of x^{y} along the line x = 0 is 0, but the limit along the line y = 0 is 1, not 0. For consistency and usefulness, a "natural" choice would be to define 0^{0} = 1.
Editor's note: This article was published in March of 2008.