#### Answer

$i$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
i^{-27}
,$ 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}
i^{-28+1}
\\\\=
i^{-28}\cdot i^1
\\\\=
i^{-28}\cdot i
.\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{1}{i^{28}}\cdot i
.\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})^{14}}\cdot i
.\end{array}
Since $i^2=-1,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{1}{(-1)^{14}}\cdot i
\\\\=
\dfrac{1}{1}\cdot i
\\\\=
1\cdot i
\\\\=
i
.\end{array}