Is $i^i$ mathematically valid? WARNING: SLIGHT NSFW

http://www.smbc-comics.com/index.php?db=comics&amp;id=2934#comic

Uhh...guys, mathematically speaking, how accurate is this comic. From what I remember in High School

$$a^b= \underbrace{\,a\cdot a\cdot  \ldots \cdot a \,}_{b\ \text{ times}} $$
So does that mean that 
$$i^i= \underbrace{\,i\cdot i\cdot  \ldots \cdot i \,}_{i\ \text{ times}} $$

How can you multiply a number by itself an imaginary number of times?
To piggy back off of @Crostul and @dbanet, the answer to the question is that the comic is mathematically accurate. One could even ask does real exponentiation make sense? If you have something like $3^{\pi}$, is that really $$\underbrace{3\cdot 3\cdot\dots \cdot 3}_{\pi \text{ times}}$$How do you multiply something $\pi$ times? Even simpler, how do you even multiply something a rational number of times? The answer is actually not too hard: Rational exponentiation makes some intuitive sense, in that $3^{2/5}=\sqrt[5]{3^2}$, or it is the number such that its 5th power is $3^2$. Then, real exponentiation can be defined to be a limit or supremum of powers of all rational numbers less than the real number. Complex exponentiation requires Euler's formula to make any sort of sense. 

For a good reference which, if you've at least seen calculus before should be readable (though maybe difficult), would be Needham's Visual Complex Analysis. It has a great intuitive explanation of Euler's formula.
No, you have to use Euler's formula for imaginary exponents.
In fact, as stated here, 
$\,i^i = e^{-\pi/2}.\,$

Indeed, on complex plane point ${\left(0,i\right)}{}$ corresponds to the angle $\,\theta = \dfrac{\pi}{2}.\,$ Therefore $\,i = e^{\,i\pi/2},\,$ and 
$$
i^{\,i} = \left(e^{\,i\pi/2}\right)^i = e^{\,i^2\pi/2} = e^{-\pi/2}
$$