Answer
$i$
Work Step by Step
Note that
$i^{17} = i^{16+1}$
RECALL:
(1) $a^{mn} = (a^m)^n$
(2) $i^2=-1$
(3) $a^{m+n} = a^m \cdot a^n$
Use rule (1) above to obtain:
$=(i^2)^{8} \cdot i$
Use rule (2) to obtain:
$=(-1)^{8} \cdot i$
Note that when $-1$ raised to an even power, the result is $1$.
Thus, the expression above simplifies to:
$= 1 \cdot i
\\=i$
