#### Answer

$-i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\dfrac{1}{i^{17}}
,$ use the laws of exponents and the equivalence $i^2=-1.$
$\bf{\text{Solution Details:}}$
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{i^{18-1}}
\\\\=
\dfrac{1}{i^{18}\cdot i^{-1}}
.\end{array}
Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{i^{2\cdot9}\cdot i^{-1}}
\\\\=
\dfrac{1}{(i^2)^9\cdot i^{-1}}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{i}{(i^2)^9}
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{i}{(-1)^9}
\\\\=
\dfrac{i}{-1}
\\\\=
-i
.\end{array}