Wikipedia User:Sainobermung

We'll give several arguments to show that the answer "should" be 1.

The alternating sum of binomial coefficients from the n-th row of Pascal's triangle is what you obtain by expanding (1-1)n using the binomial theorem, i.e., 0n. But the alternating sum of the entries of every row except the top row is 0, since 0k=0 for all k greater than 1. But the top row of Pascal's triangle contains a single 1, so its alternating sum is 1, which supports the notion that (1-1)0=00 if it were defined, should be 1.

The limit of xx as x tends to zero (from the right) is 1. In other words, if we want the xx function to be right continuous at 0, we should define it to be 1.

The expression mn is the product of m with itself n times. Thus m0, the "empty product", should be 1 (no matter what m is).

Another way to view the expression mn is as the number of ways to map an n-element set to an m-element set. For instance, there are 9 ways to map a 2-element set to a 3-element set. There are NO ways to map a 2-element set to the empty set (hence 02=0). However, there is exactly one way to map the empty set to itself: use the identity map! Hence 00=1.

Here's an aesthetic reason. A power series is often compactly expressed as SUMn=0 to INFINITY an (x-c)n. We desire this expression to evaluate to a0 when x=c, but the n=0 term in the above expression is problematic at x=c. This can be fixed by separating the a0 term (not as nice) or by defining 00=1.