The solution is:
n
0 = n
x - x
= (n
x) (n
-x)
= (n
x) (n
x)
-1
= (n
x) (1/((n
x) ))
= 1
Extra Credit:
I accepted two answers for this.
0
0 = undefined or 0
0 = 1.
Davit Banana gives his explanation:
If n = 0,
0
0 = 0
x - x
= (0
x) (0
-x)
= (0
x) (0
x)
-1
= (0
x) (1/((0
x) ))
= (0
x) / (0
x)
= 0 / 0
= undefined since division by zero is undefined.
However, Colin Bowey pointed out that Euler gives the value as 1:
0
0 = (a - a)
(n-n) = (a-a)
n / (a-a)
n = 1 (Euler)
https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero
Some authors define 0
0 as 1 because it simplifies many theorem statements in mathematics.
We define 0! = 1 for the same reason. It makes everything work right.
K. Sengupta sent in arguments for both answers:
We observe that a^0 is always 1, and accordingly 0^0 is 1
However, we also observe that 0^a is zero, whenever a>0, giving 0^0 as 0
0 is of course a real number. In view of the foregoing, it follows that a real number exponent of a real number has two distinct values, viz. 0 and 1
This is a contradiction.
Therefore, there is no agreement upon the value of 0^0, and the quantity is deemed as INDETERMINATE.
Counter argument
-------------------------------
In higher mathematics, by defining 0^0 as 1 allows some formulas to be expressed in a facile manner. Also,
(i) The interpretation of b^0 as an empty product assigns the value 1.
(ii) The combinatorial interpretation of b^0 is the number of 0-tuples of elements from a b-tuple of elements from a b-element set; there is only one 0-tuple.
(iii) The set theoretic interpretation of b^0 is the number of functions from the empty set to a b-element set; there is exactly one such function, namely the empty function.
Ref: WIKIPEDIA
So, if asked for a definitive single value of 0^0, the answer would be 1.