
The definition and domain of exponentiation has changed several times over the years. The original operation x^{y} was only defined when y was a positive integer. The domain of the operation of exponentiation has been extended for the same reason George Leigh Mallory climbed Mount Everest: "Because it's there", and also for the useful results provided by these extensions.
The original definition of exponentiation is this: x real, y
positive integer, x^{y} = 1 * x * ... * x,
where 1 is multiplied by x, y times. This operation has a number of properties, including
Now, we can try to see how far we can extend the domain of
exponentiation so that the above properties (and others) still hold. This
naturally leads to defining the operation x^{y} on the domain x positive
real; y rational, by setting x^{p/q} = the q^{th}
root of x^{p}. This
operation agrees with the original definition of exponentiation on their
common domain, and also satisfies (1), (2) and (3). In fact, it is the
unique operation on this domain that does so. This operation also has
some other properties:
Again, we can again see how far we can extend the domain of exponentiation while still preserving properties (1)(5). This leads naturally to the following definition of x^{y} on the domain x positive real; y real:
If x>1, x^{y} is defined to be sup_{q}{x^{q}} , where q runs over a ll rationals less than or equal to y.
If x<1, x^{y} is defined to be inf_{q}{x^{q}} , where q runs over a ll rationals bigger than or equal to y.
If x=1, x^{y} is defined to be 1.
Again, this operation satisfies (1)(5), and is in fact the only operation on this domain to do so.
The next extension is somewhat more complicated. As can be proved using the methods of calculus or combinatorics, if we define e to be the number
e = 1 + 1/1! + 1/2! + 1/3! + ... = 2.71828...
it turns out that for every real number x,
e^{x} = 1 + x/1! + x^{2}/2! + x^{3}/3! + ...
This series always converges regardless of the value of x.
e^{x} is also denoted exp(x).
One can also define an operation ln(x) on the positive reals, which is the inverse of the operation of exponentiation by e. In other words, exp(ln(x)) = x for all positive x. Moreover,
exp(z) = 1 + z/1! + z^{2}/2! + z^{3}/3! + ...
for all complex z (not just the reals, as before), and to define
x^{z} = exp(z ln(x))
when x is a positive real and z is complex.
This is the only operation x^{y} on the domain x positive real, y complex which satisfies all of (1)(7). Because of this and other reasons, it is accepted as the modern definition of exponentiation.
From the identities
sin(x) = x  x^{3}/3! + z^{5}/5!  z^{7}/7! + ...
cos(x) = 1  x^{2}/2! + z^{4}/4!  z^{6}/6! + ...
which are the Taylor series expansion of the trigonometric sine and cosine functions respectively. From this, one sees that, for any real x,
exp(ix) = cos(x) + i sin(x)
Thus, we get Euler's famous formula
e^{π}^{ i} = 1
and
e^{2}^{π}^{ i} = e^{0} = 1
One can also obtain the classical addition formulae for sine and cosine from (8) and (1).
All of the above extensions have been restricted to a positive real for the base. When the base x is not a positive real, it is not as clearcut how to extend the definition of exponentiation. For example, (1)^{1/2} could well be i or i, (1)^{1/}^{3} could be 1, 1/2+(sqrt(3)/2)i, or 1/2(sqrt(3)/2)i, and so on. Some values of x and y give infinitely many candidates for x^{y}, all equally plausible. And x=0 has its own special problems. These problems can all be traced to the fact that the exp function is not injective on the complex plane, so that ln is not well defined outside the real line. There are ways around these difficulties (defining branches of the logarithm, for example), but we shall not go into this here.
Some people say 0^{0} is not defined, while others say it is 1. Defining 0^{0} as 1 is useful because it gives correct answers to formulas like the number of ordered ntuples that can be made with m symbols. The formula is m^{n}. If m and n are both zero the formula gives the answer: one. This is right. The number of 0tuples that can be made with 0 symbols is one. It is ( ). Interestingly, the number of 1tuples, 2tuples or any other tuples that can be made with 0 symbols is zero  there is no way to represent a 3tuple, for example, without using symbols. Thus the formula gives correct answers for all m and n that are nonnegative.
The operation of exponentiation has also been extended to other systems like matrices and operators. The key is to define an exponential function by (6) and work from there.
http://db.uwaterloo.ca/~alopezo/mathfaq/mathfaq.html  Frequently Asked Questions in Mathematics
Hyperbolic Functions  sinh(x) and cosh(x), which are defined in terms of exp(x)
Understanding Logarithms and Exponents
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