Simplify and write in exponential form
\[\dfrac{{{2}^{-4}}\times {{15}^{-3}}\times 625}{{{5}^{2}}\times {{10}^{-4}}}\]

Hint: As it is given in the question to convert the given equation into its exponential form by simplifying it , in general a number can be written as product of power of primes.For example 26 can be written as
     \[26={{13}^{1}}\times {{2}^{1}}\]
Here 13 and 2 are primes. So we have to convert all the numbers in the factors of prime numbers and then by using laws of exponents can be solved in an easy way.

Complete step-by-step answer:
In the above given equation it is evident that 2 and 5 are already primes so now lets convert 15, 625 and 10 as product of primes
     \[\begin{align}
  & 15=3\times 5 \\
 & 625=5\times 5\times 5\times 5={{5}^{4}} \\
 & 10=5\times 2 \\
\end{align}\]
Now the given equation can be written as
     \[\begin{align}
  & \dfrac{{{2}^{-4}}\times {{\left( 5\times 3 \right)}^{-3}}\times {{\left( 5 \right)}^{4}}}{{{5}^{2}}\times {{\left( 5\times 2 \right)}^{-4}}} \\
 & =\dfrac{{{2}^{-4}}\times {{5}^{-3}}\times {{3}^{-3}}\times {{5}^{4}}}{{{5}^{2}}\times {{5}^{-4}}\times {{2}^{-4}}} \\
 & ={{5}^{3}}\times {{3}^{-3}} \\
\end{align}\]
The exponential form of the given equation is
     \[{{5}^{3}}\times {{3}^{-3}}\]

 If necessary it can be solved
  \[{{5}^{3}}\times {{3}^{-3}}=125\times \dfrac{1}{{{3}^{3}}}=\dfrac{125}{27}\]

Note: It would be easy to solve the problem if before solving the question we observe it and identify the already present prime factors. The other way to solve this problem is just by converting it into a way of representing repeated multiplications of the same number by writing the number as a base with the number of repeats.