Why is the absolute value of "$i$" one? It makes all good sense to say that $|i|=1$, but how do I justify that statement?  The equation for the absolute value of a complex number, $|a+bi|=\sqrt{a^2+b^2}$, sort of relies on $|i|=1$ or else it obviously wouldn't work.
If we define absolute value as the distance from $0$, how did we know $|i|=1$ to begin with when drawing the complex plane?  Without that, one would be unable to correctly scale the imaginary axis.
Is there a proof on why $|i|=1$?
 A: The absolute value of the complex number $z$ is defined as $|z|=\sqrt{z\bar{z}}$, after having showed that $z\bar{z}$ is a nonnegative real number: indeed, if $z=a+bi$, then $z\bar{z}=a^2+b^2$, as required.
This amounts to saying $|a+bi|=\sqrt{a^2+b^2}$ and, in particular, $|i|=1$.
A: The geometry of the Cartesian coordinate plane can be thought of as a logical prerequisite for the construction of the complex numbers. The addition and multiplication of complex numbers may be thought of as additional structures that are imposed on top of the Cartesian plane. (Historically, as @egreg points out, complex numbers appeared in the 1500s before Cartesian coordinates in the 1600s. The planar representation of a complex number appeared only in the 1700-1800s).
In a formal sense, a complex number is, by definition, an ordered pair of real numbers $(a,b)$. Also, complex addition is, by definition, just coordinate-wise addition
$$(a,b) + (c,d) = (a+c,b+d)
$$
and complex mulitplication is, by definition, given by the formula
$$(a,b) \cdot (c,d) = (ac-bd,ad+bc)
$$
Now you can feel free to assign special names such as
$$i = (0,1)
$$
and to introduce special shorthand abuses of notation such as
$$s = (s,0), \,\, s \in \mathbb{R}
$$
and using that to derive other shorthand notations such as
$$a + bi = (a,b), \,\, a,b \in \mathbb{R}
$$
and to define special geometric measurements such as
$$|(a,b)| = \sqrt{a^2 + b^2}
$$
and you are ready to prove that $i^2=-1$ and $|i|=1$.
A: First way: writing $i$ in algebraic form, i.e. as $a+ib$;
$$
i=0+i1
$$
now you know that the absolute value of a complex number $z=a+ib$ is defined as
$$
|z|:=\sqrt{a^2+b^2}
$$
thus, since for $i$ is $a=0$ and $b=1$, you get $|i|=\sqrt{0^2+1^2}=1$.
Second way: writing $i$ in polar form, i.e. as $re^{i\theta}=r(\cos\theta+i\sin\theta)$:
$$
i=1e^{i\frac{\pi}{2}}=1\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)
$$
from which you get
$$
|i|=|1e^{i\frac{\pi}{2}}|=\left|\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)\right|
=\sqrt{\cos^2\frac{\pi}{2}+\sin^2\frac{\pi}{2}}
=\sqrt{1}=1
$$
Third way: geometrical view.
The field of the complex number $\Bbb C\simeq\Bbb R^2$, is a two-dimensional vector space over $\Bbb R$; thus seeing its elements as vectors of the real plane, the meaning of the modulus of a complex number, can be seen as the lenght of the correspondant vector in $\Bbb R^2$.
Now, in the isomorphism $\Bbb C\simeq\Bbb R^2$ given by $a+ib\mapsto(a,b)\in\Bbb R^2$, $i$ is clearly the couple $(0,1)$, which is a vector of unitary lenght.  
A: After reading egreg's answer, I had an idea:
$$|a\cdot b|=|a|\cdot|b|$$
$$|i\cdot i|=|i|\cdot|i|\implies|i^2|=|i|^2$$
$$\implies|-1|=|i|^2$$
$$1=|i|^2$$
$$\implies|i|=1$$
Any reasons why this could be wrong or if this is a completely valid way of proving?
A: I don't see the need in proving it, because this holds sort-of "by definition". However, here's a proof where I use everything except for the definition of the norm. I think this is a "joke-proof", because simply assuming everything holds except for the part that is needed to prove is trivial.
Suppose $x\in\mathbb{C}$, then $x = |x|e^{i \arg (x)}$ is the polar representation of the number. Let $x=i$ and use $\arg(i) = \frac{\pi}{2}$. We know that $e^{i \arg(i)} = i\sin(\arg(i))+ \cos(\arg(i)) = i\sin(\pi/2) + \cos(\pi/2) = i$.
Substitute into the definition $x = |x|e^{i \arg(x)}$ to find $i = |i| i$. Divide by $i$ (or multiply with $-i$) to see that $|i| = 1$.
A: Here's one way we can motivate it. Assume the definition that $i = \sqrt{-1}$ (from your comment), that is to say, $i^2 = -1$. 
Now, we also want to define absolute value in a way that it extends and retains certain properties of real numbers, for example, that $|xy| = |x||y|$. So we would like it to be the case that $|i^2| = |i \cdot i| = |i||i|$. 
But looking at the initial definition, $i^2 = -1$, the absolute values of the left and right sides must be equal, and we know in real numbers that $|-1| = 1$. So the only way that $|i^2| = |i||i|$ could possibly be equivalent is to have $|i| = 1$; any other value of $|i|$ would give a different product and hence a contradiction. This then is part of what motivates the standard definition for the absolute value of complex numbers. 
