>Please can anyone explain to me why, when any real number is
>raised to the power of zero, the result is 1.

As Alex noted, 0^0 is usually left undefined, but you are correct, a^0 = 1 for any real number a not equal to zero.

Alex gave the practical reason: it is that way so that the rule a^(b_1)*a^(b_2) = a^(b_1 + b_2) holds. But if you go deeper into the definition of exponentiation, it also holds.

Defining how to raise a real number to a nonzero integer is easy, and from there it is a short step to raising any real number to any nonzero rational number. But then there is the question, what is a^b where a is any positive real and b is any real number or zero?

There are two directions that you can take in defining exponents of this form, and both are commonly used. The first is for a real number b, consider any sequence of numbers b_1,b_2,... that converge to b. Then let
a^b := lim_{n goes to infinity} a^(b_n).
The only tricky part here is showing that this definition makes sense: that is, for *any* sequence that converges to b, that you get the same limit. In fact this is true: any sequence that converges to b will have the same limiting value of a^(b_n).

For the zero case then a^0 = lim_{n goes to infinity} a^(1/n) which is just 1.

The second method used to define exponentials first defines what e^x is for all real x as the sum of a convergent series:
e^x := 1 + x + x^2/2 + x^3/3! + ...
This function e^x can be shown to be increasing, so the inverse exists and the natural logarithm (written ln(x)) is this inverse. Now exponentials can be defined:
a^b := e^(b*ln(a))
No matter what a is, when b = 0, b*ln(a) = 0 and a^0 = e^0. But from the definition of e^x:
e^0 = 1 + 0 + 0^2/2 + 0^3/3! + ... = 1.
Hence a^0 = 1. The advantage of this definition of exponentiation over the first is that (while less intuitive) it extends in a natural way to deal with negative numbers raised to real exponents.