Simplify exp[i pi sin(1/i ln(i))]

I know the answer is -1 but im not too sure how to get there?

image of it in natural form> http://imageshack.us/photo/my-images/195/picfd.png/

exp [iπ sin((1/i) log i)]
= exp [iπ sin((1/i) (ln |i| + i arg i))], definition of complex log
= exp [iπ sin((1/i) (ln 1 + i (π/2 + 2πk))] for any integer k
= exp [iπ sin(π/2 + 2πk)]
= exp [iπ * 1]
= cos π + i sin π, using complex exponential definition
= -1.

I hope this helps!