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Try looking at division rather than multiplication.

Example; 9^3 = 9x9x9. Divide this by 9, leaves you with 9x9 = 9^2. Divide by nine again leaves you with 9 = 9^1. Notice that the index is decreasing by 1 each time. So divide by 9 again, to arrive 1 = 9^0

Let A be a number other than zero. A^1 = A, A^2 = A*A. What about (A^1)(A^2)? Well that's A*A*A = A^3. So all we had to do was add the exponents to get the right answer.

Now what about A^(-1)? That's the same as 1/A. Similarly, A^(-2) = 1/(A^2).

Now what about (A^2)(A^(-2))? We can write that as A^2/A^2. And of course any number divided by itself is 1.

If we add the exponents we get A^(2-2) = A^0. Therefore A^0 must be 1 as long as A is not zero.

Let "a" be any real number (-∞, ∞).

Therefore, we can say that:

1 = (a^n) / (a^n)

Now let's go ahead and re-write this as:

1 = (a^n)*(a^-n)

(a^n)*(a^-n) = a^0

Therefore:

1 = a^0

In short:

1 = (a^n) / (a^n) = (a^n)*(a^-n) = a^0

And as a result, any number raised to the 0 power will always equal 1.

One way to look at this:

we assume x^a*x^b=x^(a+b) for all real a,b.

Setting b=0 yields

x^a*x^0=x^a=>x^0=1.

How you understand this depends very much on how properties

of exponentiation have been developed for you. There is more than one way.

take a variable x

now x/x=1 ......simple division

x can be written as x=x^1

and 1/x can be written as x^-1

x/x=x*(1/x)

=x^1*x^-1

=x^(1-1)

=x^0

we knowx/x=1

hence x^0=1

also try this from logarithms

log of 1 to any base is 0

in equations,....(log1)base n =0

taking out logs from both sides 1=n^0

or simply any number n raise to zeroth power is 1

consider x^a/x^b then by exponents this is x^[a-b]

now take a = b then original expressionion is x^a/x^a =1 and the second exp x^0

which must be the same so x^0=1

something really hard to explain, and it makes sense just when you really understand the Exp function. remember e^1, so it is just a constant value. but it is kind of hard.