2^5 =32

2^4 =16

2^3 =8

2^2 =4

2^1 =2

2^0 =1

2^-1 =1/2

2^-1 =1/4 and so on

Notice that the answers decrease by dividing the previous answer by the base number. ( or vice-versa, they increase by multiplying the prior answer by the base.) No matter what the base is you will always base to the 1st power equal to the base itself . By dividing by the base value to get the zero power you will always get 1. Hope this helps.

n^x

— = n^(x-y)

n^y

for all n, x, and y. So for example,

3^4

— = 3^(4-2) = 3^2

3^2

3^4

— = 3^(4-3) = 3^1

3^3

Now suppose we have the fraction:

3^4

—

3^4

This fraction equals 1, because the numerator and the denominator are the same. If we apply the law of exponents, we get:

3^4

1 = — = 3^(4-4) = 3^0

3^4

So 3^0 = 1.

We can plug in any in number in the place of three, and that number raised to the zero power will still be 1. In fact, the whole proof works if we just plug in x for 3:

x^4

x^0 = x^(4-4) = — = 1

x^4

Putting in formula it is 2^3/2^3. which is 8/8 any number divided by itself is 1.

The extension to the formula is n^(x-y) =n^x/n^y you got n^x/n^y=1 which must equal n^(x-y) so

2^(3-3)=1 which is 2^0=1

Hence all number with power of 0 =1

So a^n divided by a^n = a^(n-n) = a^0

But a number divided by itself is 1. So a^n / a^n = 1

Therefore a^0 = 1.

That is unless a is zero. Then it’s undefined

RE:
Please explain why a number to the zero power is always 1?