How do you rewrite $\dfrac{1}{{{a}^{4}}}$ with negative 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. We find the relation between negative exponent and inverse of the number to find the solution.

Complete step by step answer:
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 ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
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, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
We know that ${{a}^{0}}=1$. So, we replace the value of 1 in the numerator in the $\dfrac{1}{{{a}^{4}}}$.
$\dfrac{1}{{{a}^{4}}}=\dfrac{{{a}^{0}}}{{{a}^{4}}}={{a}^{0-4}}={{a}^{-4}}$.
In case of inverse exponent, we just take the negative value of the indices.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices. For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$. The relation is independent of the values of $m$ and $n$.