To carry out this proof, I shall do the following sequence of proofs:
The caret symbol, ^, means "raised to the power of".
An expression |z|^n means that the number |z| is multiplied by itself n times. Therfore if n is increased by 1 a larger number |z| * |z|^n results. Futher increases of the exponent n will add new multipliers to the product. One multiplier, |z|, will be added to the product for each unit increase in the exponent so that larger exponents will create larger products. Also there will be one multiplier less for each unit decrease in the exponent giving smaller products for smaller exponents. Then, the exponential product of the absolute value of a radix grows and shrinks in magnitude in proportion to the magnitude of the exponent. The product is monotonic. This is proven for positive integer exponents.
|z|^n is a product of |z| times itself n times. |z|^m is a product of |z| times itself m times. Then the product of |z|^n * |z|^m is a product containing the products of n multipliers |z| and m multiplicands |z|. The combined number of factors |z| contained in the product |z|^n * |z|^m is equal to n + m which is also the sum of the exponents. Therefore exponents of a radix combine through the combination of numbers of copies of the radix in each factor of the product. That is, counting the number of copies of the radix in the new product and adding the exponents of each factor with the radix will give the same number for positive integer exponents.
We define |z|^1 as equal to |z|. We also presume that the addition of exponents in products of powers of a radix applies to positive fraction exponents as well as integer exponents. Then any product of powers of a radix, |z|, whose sum of exponents equals one must be derived from the sum of fractional exponents. These same fractional powers of |z| when multiplied must equal |z|. So |z| = |z|^[n*1/n] = [|z|^(1/n)]^n is a reprentation of the production of |z|^(1/n) with itself n times so that the fractions in the exponent adds to equal 1. The addition of 1/n, n times, is replaced by the multiplication [n*1/n], both of which clearly equals 1. It can be shown through the properties of roots, namely [nth_root(|z|)]^n = nth_root(|z|^n) = |z|, that the production of the nth_root of |z|, n times, will also equal |z|. Therfore we are allowed to write that if [|z|^(1/n)]^n =[nth_root(|z|)]^n then |z|^(1/n) = nth_root(|z|). That is, the fractional powers are equivalent to the roots of the radix.
We extend the proof of the monotonic property of powers of a radix |z| to include positive fractions. This is possible through the previous proof where smaller roots have smaller magnitude that larger roots. Then the limit of roots as the fracrional power gets infintesimally smaller is zero. Now it is defined that 1/n equals 0 as n becomes infinity. To reasonably satify [|z|^(1/n)] =[nth_root(|z|)], we assume that there is no common number that can be muliplied times itself an infinite number of times and not be equal to infinity except 0 and 1. It is easy to show graphically that as one plots the roots of the radix |z| that the graph will clearly approach 1 and not zero. Also if we were to presuppose that it is true that the negative exponents are to be writen as fractions in the power of the radix then this current proof would be easily demonstrated by showing that |z|^n * |z|^(-n) = z^0 by the addition of exponents and |z|^n * |z|^(-n) = (|z|^n )/(|z|^(n)) =1 by the inversion of negative powers, then |z|^0=1. At this point we can not use this result because |z|^0=1 is part of the proof of the negative exponent conjecture and must be proved first. We must either accept the graphical proof or prove this present conjecture symbolically in order to preserve the rigor of this proof. To do this symbolically our best chance is to show that 0 cannot be a viable solution of |z|^(1/inf.) = inf._root(|z|) = x. If x = 0 then |z| must have been zero and x would be zero for all n. But we know the |z| is not constained to be zero always therefore 0 is not a viable solution. Further, it can be shown numerically that the nth_roots of every non-zero positive number are greater than 1 and approaches 1 as n approaches infinity; therefore x =1 is a viable solution. Exact symbolical proof leads one into calculus and indeterminate forms that also depend upon the negative exponent conjecture and so it will not be prosecuted here. In general, it is difficult to prove that |z|^0 = 1 without proving the negative exponent conjecture as it is difficult to prove the negative exponent conjecture without proving that |z|^0 = 1.
To show that negative exponents are indeed the representation of fractions in powers of the radix let |z| be any radix and n be any exponent. For simplicity assume that both |z| and n are integers but the proof is equally valid if they are real numbers.
It was shown above |z|^n is greater than |z|^(n-1), which in turn is greater than |z|^(n-2) for some integer n. This sequence can continue until |z|^(n-n)=z^0=1 is reached. The sequence that is established by the monotonically decreasing powers of z is such that the magnitude of every value of a power is less than the magnitude of the previous value.
Now, when the power reaches zero, there are no more positive integers. In order for the sequence to continue any further, exponents must be less than zero or negative. Also, the value of the power which follows |z|^(n-n)=|z|^0=1 is |z|^(n-n-1)=|z|^(-1) and must be less than its predecessor and therefore must be a fraction. Also, it must be a power of |z| to belong to the sequence and continue the pattern established with the positive exponents. The only number that satifies both of these criteria is the fraction 1/|z|^1. Likewise |z|^(-2) must be written as the fraction 1/|z|^2 to be a member of the sequence of decreasing powers of |z|.
The negative exponents, therefore, generate fractions as a consequence of the monotonic nature of the powers of a radix and the pattern established in the sequence generated by monotonically decreasing exponents of that radix. Q.E.D.