Why is it that any number raised to the power of zero is always one and not zero.

look at 2^2=4 2^3=8 the pattern is that you multiply by 2 each time
well 2^1=2 divide by two to get 2^0 so 2/2=1

By the way , it is NOT ANY number.....It is any number EXCEPT 0 raised to the 0 power

Using the exponent laws:::::::::::

you agree (x^n)/(x^n)= 1.................7/7=1..........x/x=1(B… x NOT equal to 0 since you dont divide by 0)

Exponent wise (x^n)/(x^n)= x^(n-n)=x^0

By the transitive property (If a = b and b=c, then a=c),

1=(x^n)/(x^n)= x^0,

Thus 1 = x^0, as long as x does not equal 0

The short answer is that a^0 has been defined as one.

Since I personally don't like "by definition" answers, here's a slightly longer explanation:

Remembering that a^1 = a, and that a^(n-1) = (a^n) / n, if you set n = 1
then
a^0 = (a^1) / a
= a / a
= 1

Let us take a number x^0
Now we can rearrange it as
x^0 = x^(1-1) [Or any general format like x^(n-n) where n could be any number]
=> x^0 = x^1 * x^-1
=> x^0 = x/x
=> x^0 = 1

this is true for any value of x

Because the Laws of Exponents are followed.
Like 2^0=1
1^0=1
3^0=1

5^2 * 5^3
= 5^(2+3)
= 5^5

5^2 * 5^3 * 5^0
= 5^(2+3+0)
= 5^5

If both equations are the same, then 5^0 has to equal 1, not zero.

hi there,
if you are familiar with exponent
2^1/ 2^1 = 2^(1-1) = 2^0
this is the same as
2^1/ 2^1 = 2/ 2= 1
which means 2^0 = 1