Maxwell predicted it as he did with Saturns rings being made of small
pieces.
Quote from Wikipedia:
The Faraday effect, discovered by Michael Faraday in 1845, was the first experimental evidence that light and electromagnetism are related.The
ability to rotate polarized light in a magnetic field showed light to be
an electromagnetic wave.

Charged particle motion state changes can generate or absorb photons, electronic, for example: electronic excitation energy is increased to at least three ways: 1) the thermal motion of molecules or atoms collide with each other makes, then it may will be part of the kinetic energy of molecules or atoms transferred to the electron in the collision, and then return it when the low-level light. Incandescent lamp is a typical example. 2) absorption of the incident photon and electron excitation, return to the low level, may not return original way, but the first jump to a lower energy level, then jump to a lower energy level, so that we can send more multiple smaller photon. Mercury atoms emit fluorescent light in the ultraviolet photons are absorbed by the wall elements on the phosphor to emit visible light after a typical example. 3) an alternating electric field can promote the free electrons in the conductor to do swing movement - oscillating current in the circuit up, so you can send radio waves.

There are 4 basic equations in Electromagnetism called "Maxwell's equations":

(1) Div E=ρ/ε (2)Curl E=-∂B/∂t (3)Div B=0 (4)Curl B=μJ+με ∂E/∂t

These are partial differential equations and once you have some thing called the boundary condition, you basically can expressed everything within the study of electromagnetism.

One uses the theorems of vector analysis to derive 2 equations from these equations:
(5) ∆^2E=εμ ∂^2E/∂t^2 and (6)∆^2B=εμ ∂^2B/∂t^2
which matched something called the "wave equation", that expresses for example waves in sounds:

(7)∆^2A=1/(v^2) ∂^2A/∂t^2

where v is the speed of the wave.

if you compare for example (5) and (7) you can see that 1/v^2=εμ......so v=Square root of (εμ).
the quantity of ε and μ were already found emperacly and the square root of the product of thoes 2 became exactly the same as the speed of light.

1: Take the Curl on both sides of equation (2) : Curl(Curl E)=-∂Curl(B)/∂t
here Curl(Curl E)=Grad(Div E)-∆^2E=-∆^2E (Since Div E is 0).
Hence
∆^2E=∂Curl(B)/∂t =εν∂^2 E/∂t^2
(from using also equation (4) assuming J=0);

2:Take the Curl of (4) and do a similar procedure, you will get:
∆^2B=εμ ∂^2B/∂t^2