Answer
$-i$
Work Step by Step
Note that:
$i^{15} = 1^{14+1}$
RECALL:
(1) $a^{mn} = (a^m)^n$
(2) $i^2=-1$
(3) $a^{m+n} = a^m \cdot a^n$
Use rule (3) above to obtain:
$=i^{14+1}=i^{14} \cdot i$
Use rule (1) above to obtain:
$=(i^2)^7\cdot i$
Use rule (2) to obtain:
$=(-1)^{7} \cdot i
\\=(-1)(-1)(-1)(-1)(-1)(-1)(-1)(i)
\\=-i$
