How do you simplify ${{\left( 2{{e}^{-2}} \right)}^{-4}}$ ?

Hint: This question can be solved by using the law of exponents. Laws of exponents are as follows:
Product of powers: ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
Quotient of powers: ${{a}^{m}}\div {{a}^{n}}={{a}^{m-n}}$
Power of a power: ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$
Power of a product: ${{\left( xy \right)}^{m}}={{x}^{m}}\times {{y}^{m}}$
Power of a quotient: ${{\left( \dfrac{x}{y} \right)}^{m}}=\dfrac{{{x}^{m}}}{{{y}^{m}}}$
Negative exponent: ${{x}^{-m}}=\dfrac{1}{{{x}^{m}}}$ or ${{x}^{m}}=\dfrac{1}{{{x}^{-m}}}$
Identity exponent: ${{x}^{1}}=x$
Zero exponent: ${{x}^{0}}=1(x\ne 0)$
We need to write the final answer in exponential form only because our base is an exponent (e) itself and not a number here. The question is given in the natural base so the answer should also be in a natural base.

Complete step by step answer:
Now, let’s simplify the given expression in the question.
First, write the expression given in the question:
$\Rightarrow {{\left( 2{{e}^{-2}} \right)}^{-4}}$
Now in the second step, we have to apply a negative exponent or multiplicative inverse to this expression in order to make the power positive.
$\Rightarrow \dfrac{1}{{{\left( 2{{e}^{-2}} \right)}^{4}}}$
In the third step we have to multiply 4 with the powers inside.
$\Rightarrow \dfrac{1}{{{2}^{4}}\times {{e}^{-2\times 4}}}$
On further solving:
$\Rightarrow \dfrac{1}{16{{e}^{-8}}}$
Now, once again we have to apply a negative exponent or multiplicative inverse to this expression in order to make the power positive.
$\begin{align}
  & \Rightarrow \dfrac{1}{\dfrac{16}{{{e}^{8}}}} \\
 & \Rightarrow \dfrac{{{e}^{8}}}{16} \\
\end{align}$
Leave the answer in exponential form only as the base is natural base i.e. (e).

Note: Most of the children consider that power of a power rule is applied in the starting which is a wrong start of this question. Before solving such types of questions, you should know all the laws of exponents.