Would someone like to enlighten me on how exactly the speed of light came into be in this equation? With all the power and experimental confirmation this equation has, I am completely clueless to how it was derived. Any help is appreciated. :D
There's a short and a long answer to your question. One is rather mathematical, the other is rather intuitive. I am not sure if this is the way the equation got derived by Einstein, but I looked it up in three textbooks and they more or less introduced the equation the same way:
Short answer: In studying the momentum of a free particle in the special theory of relativity, suddenly a number appears with the dimension of energy. When explicitly evaluated, this number gives mc^2. Physicists just take this to be the energy of the particle.
Long answer: Physicists study the dynamics of free particles in a mathematical construction called the Minkowski space (this is a four dimensional space that unifies time and space in one description, in contrast to the Newtonian way of describing space in a separate three dimensional Euclidian space and time as a one dimensional parameter). Since this Minkowski space is 4-dimensional, the "spacetime" location of a particle is described by what is called a 4-vector: this vector contains time and the three spatial coordinates of a particle. The speed of light is very important in this Minkowski space: it is unnatural to express the 4-vector in 2 different units: seconds (for time) and meters (for space), so instead of writing (t,x,y,z) as a coordinate, physicists write (c*t,x,y,z). This is because c*t (the speed of light times the time) has the same dimension as space: meters. This is rather convenient, because that way you can define a scalar product on the Minkowski space; the size of a 4-vector squared is taken equal to: (c*t)^2 - x^2 - y^2 - z^2. A scalar product gives a relation between the components of 4-vectors and you want everything in this scalar product to be of the same unit: meters.
Since a 4-vector describes a location in spacetime, you can also associate a momentum with it, called the 4-momentum. Because there's a lightspeed c in the 4-vector, you'll also find this c in the expression of the 4-momentum. It then turns out that the size of the 4-momentum for a free particle is just its restmass m0*c^2. This means that for a 4-momentum P, you can write:
size squared = (first component)^2 - p^2
where p^2 is the total space momentum (which is related to the classical Newtonian momentum p = m*v for the free particle). And "first component" is the first component of the 4-momentum (which has something to do with c*t since for the location the first component was c*t). Since the size squared is the rest mass you get:
(m0*c)^2 = (first component)^2 - p^2
(first component)^2 = (m0*c)^2 + p^2
It turns out that the righthand side has a dimension (Energy/c)^2, so if we call the first component of the 4-momentum E/c, we get:
(E/c)^2 = (m0*c)^2 + p^2
Or by multiplying both sides by c^2:
E^2 = m0^2 * c^4 + p^2 * c^2
Which is just another form of the "more popular" equation E = mc^2.
This might look like some magical tricky stuff, but it makes sense if you look at it in the whole (meaning that if you work out everything nice and mathematically, that it seems to be consistent). For more information I can only suggest getting a book on special relativity and studying it (a nice, non-mathematical book is "Spacetime physics" by Taylor and Wheeler, however I personally think it's not mathematical enough and therefor sometimes obscure ;)).
Anyway, I hope this more or less answers your question.
Gotcha. Mathematical explanations do a lot more for me, so thanks for not dumbing it down TOO much. :D I understand the derivation, but one thing still troubles me: Why was the speed of light chosen to match dimensions in the Minkowski space in the first place? I may just be missing an obvious concept, but I can't seem to get it; it still seems to be an arbitrary choice. Maybe you're right, I probably should get a book. :) Anyway, thanks for the quick answer.