How do you simplify ${{i}^{7}}$?

Hint: Since $i$ is defined as the square root of negative of one, so its square root will be equal to one, that is, ${{i}^{2}}=-1$. On squaring this equation we will obtain ${{i}^{4}}=1$. On further squaring, we will obtain ${{i}^{8}}=1$. The given expression is to be multiplied and divided by $i$ to get $\dfrac{{{i}^{8}}}{i}$ which will be simplified to $\dfrac{1}{i}$ after substituting ${{i}^{8}}=1$. Then finally multiplying and dividing the numerator by $i$ and substituting ${{i}^{2}}=-1$ we will get the final simplified expression.

Complete step by step solution:
We know that $i$ is defined as
\[\Rightarrow i=\sqrt{-1}\]
Taking the square of both the sides, we get
$\Rightarrow {{i}^{2}}=-1........\left( i \right)$
Again taking square on both the sides of the above equation we get
$\Rightarrow {{i}^{4}}=1........\left( ii \right)$
Now, let us write the given expression as
$\Rightarrow E={{i}^{7}}$
Multiplying and dividing by $i$, we get
$\begin{align}
  & \Rightarrow E=\dfrac{{{i}^{7}}\times i}{i} \\
 & \Rightarrow E=\dfrac{{{i}^{8}}}{i} \\
\end{align}$
The numerator of the above expression can also be written as
$\Rightarrow E=\dfrac{{{\left( {{i}^{4}} \right)}^{2}}}{i}$
Substituting (ii) in the above expression, we get
$\begin{align}
  & \Rightarrow E=\dfrac{{{\left( 1 \right)}^{2}}}{i} \\
 & \Rightarrow E=\dfrac{1}{i} \\
\end{align}$
Again multiplying and dividing by $i$, we get
$\Rightarrow E=\dfrac{i}{{{i}^{2}}}$
Finally, substituting (i) in the above expression, we get
$\begin{align}
  & \Rightarrow E=\dfrac{i}{-1} \\
 & \Rightarrow E=-i \\
\end{align}$

Hence, the given expression is simplified as $-i$.

Note: Instead of multiplying and dividing the given expression by $i$, we can also extract the highest power of ${{i}^{2}}$ from the given expression. In doing so, we will obtain the given expression as ${{\left( {{i}^{2}} \right)}^{3}}i$. Then, as we know that ${{i}^{2}}=-1$, the expression will get reduced to ${{\left( -1 \right)}^{3}}i$ which will be simplified to $-i$ which we have obtained in the above solution.