What expression is equivalent to ${{\left( 2a \right)}^{-4}}$?

Hint: We know that, by using the property of negative powers, we can write ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. Then, by using the property of exponents, ${{\left( ab
\right)}^{m}}={{a}^{m}}\cdot {{b}^{m}}$, we can simplify our given expression into an easily understood form.

Complete step-by-step solution:
We know that a negative power on any expression is, nothing but, the reciprocal of that expression raised to the absolute value of that power.
Thus, it is clear to us that the negative power is multiplicative inverse of the base raised to the positive opposite of the power.
We can express this mathematically as follows,
${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$,
where a is any number except zero and n is any number possible, either real or imaginary.
We are also aware of the exponential property for a number raised to the power of product of two numbers, that is,
${{\left( ab \right)}^{m}}={{a}^{m}}\cdot {{b}^{m}}...\left( i \right)$
We need to find a simplified expression for ${{\left( 2a \right)}^{-4}}$.
Here, by using the equation ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$, we can easily write
${{\left( 2a \right)}^{-4}}=\dfrac{1}{{{\left( 2a \right)}^{4}}}$.
Now, let us use equation (ii) on the right hand side of this equation. Hence, we get
${{\left( 2a \right)}^{-4}}=\dfrac{1}{{{\left( 2 \right)}^{4}}\cdot {{\left( a \right)}^{4}}}$.
We know very well that ${{2}^{4}}=2\times 2\times 2\times 2=16$ and so, we can write
${{\left( 2a \right)}^{-4}}=\dfrac{1}{16{{a}^{4}}}$.
Hence, we can clearly say that the simplified form for the expression ${{\left( 2a \right)}^{-4}}$ is $\dfrac{1}{16{{a}^{4}}}$.

Note: We can also solve this problem by splitting -4 into the product of 4 and -1, that is, $-4=-1\times 4$. Thus, our expression becomes ${{\left( 2a \right)}^{-1\times 4}}$, which we can also write as ${{\left( {{\left( 2a \right)}^{-1}} \right)}^{4}}$. We know that ${{\left( 2a \right)}^{-1}}$ is nothing but $\dfrac{1}{2a}$, and so we get ${{\left( \dfrac{1}{2a} \right)}^{4}}$, which we can solve very easily.