Anything raised to the power of zero is one. Zero to any power is zero. So what's zero raised to the power of zero? It's a befuddling question.

Zero to the zeroth power, often written 0^0 in computers and more commonly seen as 0^{0} on paper and blackboards, is one of the spicier indeterminate forms because, well, it is not always indeterminate. In fact, it's hardly ever indeterminate. "Nonsense!", you protest. "If it's not always clearly equal to the same thing, it must be indeterminate." Yes, yes, yes, that's quite true, but we are speaking mathematics here, and all these words, such as "indeterminate" have special meaning. Allow me to explain.

All things being equal and with no further context, 0^{0} = 1. The reason for this is similar to the reasons why 0! = 1 and why 1 is not a prime number: they are simply conventions that make some formulae and theorems easier to state. We can of course make 0! = 3 or 0^{0} = 42 if we so want; it's just a definition, and as ariels sagely spouted, "It's a free country." But not all definitions are created equal, and some simplify matters.

Here are a few reasons why, by itself, 0^{0} should be 1:

- The binomial theorem.
We know, for example, that (*a* + *b*)^{2} = *a*^{2}*b*^{0} + 2 *a*^{1}*b*^{1} + *a*^{0}*b*^{2}. If we want this formula to still be true when *a* or *b* is 0, then we need 0^{0} = 1.

- †Analytic functions.
If functions *f* and *g* are analytic at 0, *f*(0) = *g*(0) = 0, and *f* isn't constantly equal to 0, then it is a theorem that

** g(x)
lim |f(x)| = 1,
x->0
**

where we take the absolute value of

*f*(

*x*) to avoid the unpleasantness associated with taking powers of negative numbers.

Power series.
This reason is similar to the binomial theorem reason and why 0! = 1. For example, in the exponential series

** ∞
--- n
x \ x
e = / ----,
---- n!
n=0
**

if we wish this series to be true for all values of

*x*, even 0, we need to define 0

^{0} = 1.

Intuition.
Zero to the zero power is an "empty product". There is nothing being multiplied together. Empty products should be equal to the multiplicative identity, one, just as empty sums are equal to the additive identity, zero. If we haven't multiplied, then we are at the beginning of multiplication, and multiplication starts from the unit, one.

Very well, these have been plausibility arguments as to why the definition 0^{0} = 1 is the desired one. We cannot prove that definitions are correct; they are a matter of intuition, faith, hope, and "common sense". I have said that 0^{0} is also an "indeterminate form", like 0/0, ∞/∞ or ∞ - ∞. In these cases, we are talking about limits. Here is a simple example of what can happen.

Let *a* be any positive real number less than 1, (0 < *a* < 1) and define functions *f* and *g* by

1/x
f(x) = a
g(x) = x,

for all positive *x*. Then both

lim f(x) = lim g(x) = 0.
x->0+ x->0+

So, in a sense, the limit

g(x)
lim f(x)
x->0+

is an indeterminate form 0^{0}. But we can calculate this limit. It is simply

/ 1/x \ x
lim (a ) = lim a = a,
x->0+ \ / x->0+

which is any value between zero and one, whatever we assign to *a*. In this sense, 0^{0} is indeterminate. Notice that I had to pick a slightly naughty function *f*. It is not analytic, nor even defined at zero.

To wrap up, in most situations 0^{0} is 1, except when we are dealing with limits of non-analytic functions, in which case we need to contend with indeterminate forms. Save for those few cases, there really isn't anything mysterious or contentious about multiplying zero of nothing.

___________________________________________

†This has a very special meaning. Follow the hardlink, if necessary. Analytic functions are among the most mild-mannered and polite functions we ever encounter. They deserve to be treated nicely and be given the right definitions.