How do you simplify ${{i}^{17}}$?

Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities and then we use the imaginary value and find all the conditions related to $i$. We find the simplified form of ${{i}^{17}}$.

Complete step-by-step solution:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$. In case the value of $n$ becomes negative, the value of the exponent takes its inverse value. The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
If we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
For given equation ${{i}^{17}}$, we convert it as ${{i}^{17}}={{i}^{16+1}}={{i}^{16}}\times i$.
We also have the identity of ${{a}^{mn}}={{\left( {{a}^{m}} \right)}^{n}}$. Therefore, \[{{i}^{16}}={{\left( {{i}^{4}} \right)}^{4}}\].
We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. We place the values and get
${{i}^{16}}={{\left( {{i}^{4}} \right)}^{4}}={{1}^{4}}=1$. This gives ${{i}^{17}}=1\times i=i$
Therefore, the simplified form of ${{i}^{17}}$ is $i$.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices. For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$. We need to remember that the condition for ${{a}^{m}}={{a}^{n}}\Rightarrow m=n$ is that the value of $a\ne 0,\pm 1$.