How do you simplify $\dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}$ ?

Hint: We have been given 4 terms of the same exponent function raised to 4 different powers. These have been written in the numerator and denominator and thus expressed as a fraction. We shall first analyze the positive and negative powers and then rearrange them again in the numerator and the denominator. Accordingly, we shall cancel the similar terms and write the fraction in its simplest form.

Complete step by step solution:
We know that whenever two or more like terms are multiplied, their respective powers are added and all the like terms are combined as one single term.
Similarly, whenever two or more like terms are divided, their respective powers are subtracted and all the like terms are combined as one single term. The power of the term in the numerator is subtracted from the power of the term in the denominator.
Given that, $\dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}$.
Here, we have 2 terms each in the numerator and the denominator. Thus, we shall first add the powers of the terms in the numerator and the powers of the terms in the denominator individually.
$\Rightarrow \dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}=\dfrac{{{e}^{-3+4}}}{{{e}^{-2+\left( -1 \right)}}}$
$\Rightarrow \dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}=\dfrac{{{e}^{1}}}{{{e}^{-3}}}$
Now, we shall subtract the power of the term in the denominator from the power of the term in the numerator.
$\begin{align}
  & \Rightarrow \dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}={{e}^{1-\left( -3 \right)}} \\
 & \Rightarrow \dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}={{e}^{1+3}} \\
\end{align}$
$\Rightarrow \dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}={{e}^{4}}$
Therefore, $\dfrac{{{e}^{-3}}.{{e}^{4}}}{{{e}^{-2}}.{{e}^{-1}}}$ is simplified as ${{e}^{4}}$.

Note:
Another method of solving this problem was multiplying and dividing the given term one-by-one with the lowest exponent term out of the given four terms. Doing so, we could have eliminated the denominator segment completely and then simply added all the terms in the numerator to compute our final solution.