Is $i^i$ mathematically valid? WARNING: SLIGHT NSFW
http://www.smbc-comics.com/index.php?db=comics&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?
 A: 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.
A: $a^b=\underbrace{a\cdot a\cdot\ldots\cdot a}_{b~\text{times}}$ is valid only for $b\in\mathbb N_0$. Even for real numbers like $2^{\sqrt{2}}$ you need something different, usually one defines this using exponential and logarithm: $a^b=\exp(b\ln(a))$ for $a>0$. 
When looking at $a^z$ with $a\in\mathbb C^*,z\in\mathbb C$ this gets even more complicated as we then have to define the logarithm for complex numbers first and it gets even more "complex" (pun intended) when you consider that there are different branch of logarithms one can use.
A: Well, exponentiation isn't just defined for reals. For complex numbers in particular, $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. , and for any positive $b$, $b^z = e^b\ln(z)$.
A: 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 
$$
i^{\,i} = \left(e^{\,i\pi/2}\right)^i = e^{\,i^2\pi/2} = e^{-\pi/2}
$$
