PROOF:-

As we know that;

x^1= x :- (LET it be :-1st series)

x^2= x^2 :-(LET it be :-2nd series) and,

x^3= x^3:- (LET it be :-3rd series)

Examining the above three series you can see that the product of all the three series is differ by x or you can say they have the common difference of the value x for your better understanding I am elaborating the above series:-

x ,x^2 , x^3(This are all the values of the above series and you can see that it differ by 'x' this means that the given series is in A.P)

Therefore, the value just before x will be 1 because ,

x^0 = x^(1 - 1)

x^0 = [x^1] x [x^(-1)]

x^0 = x  (1/x) [since x^1 = x and x^(-1) = 1/x]

x^0 = x / x

x^0 = 1

Therefore, X= 1 Or X^0= 1

Hence proved

Hope it helps!!

Hello,

5^0 is the product of zero 5 s, which is the empty product  , and there are plenty of reasons to define that as 1 .

We know that x^0=x×x (0 Times)

we all know 1−1=0

so we can say that x^0=x^[1−1]

Now We Got x^0=(x^1)×(x^−1)

x^0=x^1×[1/x^1]

where x^1 and x^1 gets cancelled

Now we got x^0=1

Hence Proved.