How can $i^i = e^{-\pi/2}$ !! I was asked a homework question: find $i^i$. The solution provided was as follows:

Let $A = i^i$.

$\log A = i \log i$.

Now, $\log i = \log e^{i\pi/2} = \frac{i\pi}{2}$.

So, $\log A = -\frac{\pi}{2}$

Thus, $i^i = e^{-\pi/2}$.

I understood how the result was obtained, but it is illogical. I understand that multiplying by $i$ is equivalent to rotating the position vector of the complex number in Argand Plane by $90$ degrees anti-clockwise. How can rotating $i$ anti-clockwise $i$ number of times give $e^{-\pi/2}$?

So can somebody explain to me graphically or more intuitively, how $i^i = e^{-\pi/2}$ ?
By definition $$e^{ix} = \cos x+i\sin x = cisx$$

This definition can be proved by observing the Taylor expansions of both the RHS and LHS. You will find they are both identical. 

Letting $x= \frac{\pi}{2}$ we get this: 

$$i = e^{\frac{i\pi}{2}}$$

Then just play with the powers. 

$$i^i = e^{\frac{i^2\pi}{2}}$$

$$i^2=-1$$
Hence, 
$$i^i = e^{-\frac{\pi}{2}}$$

Of course $i^i$ attains an infinite elemental set of real values (due to the periodic nature of $cisx$) but since your question was to prove $i^i = e^{-\frac{\pi}{2}}$, I have done so accordingly.
If you would like, there is also another solution to this problem. Since we cannot count &quot;i&quot; times, the solution is to find a formula for doing exponents, namely, Euler's formula. When you plug in $e^{xi}$, the output will be $\cos(x)+i\sin(x)$ (in radians), also notated as the $\operatorname{cis}(x)$.  Now, we can define $i$ as $e^{π/2i}$. So therefore our goal is $(e^{π/2i})^i$. Luckily, with the way that exponents work, we can simplify this to $e^{π/2i\cdot i}$, which of course, by definition, is simplified to
$e^{-π/2}$.