How do you simplify ${{\left( {{x}^{2}} \right)}^{0}}$ and write it using only positive exponents?

Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. Finally, we use the obvious identity theorem of ${{a}^{0}}=1$.

Complete step by step solution:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
The simplified form of the expression ${{a}^{n}}$ can be written as the multiplied form of number $a$ of n-times.
Therefore, ${{a}^{n}}=\underbrace{a\times a\times a\times ....\times a\times a}_{n-times}$.
The value of $n$ can be any number belonging to the domain of real numbers.
Similarly, the value of $a$ can be any number belonging to the domain of real numbers.
In case the value of $n$ becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
There is also an obvious rule of ${{a}^{0}}=1$ for any value of $a$.
The multiplication of these exponents works as the addition of those indices.
For example, we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. We also have \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\]
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
For our given expression, we need to express ${{\left( {{x}^{2}} \right)}^{0}}$ in its simplest form.
We use the identity of \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}\] to get ${{\left( {{x}^{2}} \right)}^{0}}={{x}^{2\times 0}}={{x}^{0}}=1$.

Therefore, the simplified form of ${{\left( {{x}^{2}} \right)}^{0}}$ is 1.

Note: We can also prove the identity of ${{a}^{0}}=1$.We can use the form of ${{a}^{0}}$ as ${{a}^{0}}={{a}^{m-m}}=\dfrac{{{a}^{m}}}{{{a}^{m}}}=1$.There is also some theoretical approach to it.
The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.