complex exponentiation: evaluating $i^{1+i}$

Question

I am trying to find evaluate the expression $i^{1+i}$.

I know that, \begin{align} i^{i+1}&=\exp((1+i)\log(i)) \ \ \ \ \text{(where $\log$ is multivalued)} \\ &=\exp((1+i)(\ln|i|+i\arg(i)+2k\pi i) \ \ \ \ \ (k\in\mathbb{Z}) \\ &=\exp((1+i)\left(i\frac{\pi}{2}+2k\pi i\right) \\ &=\exp\left(i\frac{\pi}{2}+2k\pi i-\frac{\pi}{2}-2k\pi\right) \end{align} From this step, my questions is why does $2k\pi i$ vanish? The answers I have remove it without justification.

Answer 1

1. $e^{a+b}=e^ae^b.$

2. $e^{2k \pi i}=1.$