Before we even consider this, consider the kinetic energy of a relativistic particle:
$$E=mc^{2}\left(\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}-1\right)$$
This represents the amount of energy required to accelerate the particle from rest (with respect to a given reference frame) to the speed $v$. It should be immediately obvious that this quantity becomes infinite as $v\rightarrow c. {}^{1}$ Since it is impossible to generate an infinite amount of energy, this then means that no particle with $m > 0$ can be accelerated to the speed of light. In fact, since high-energy particles have speeds that are generally extremely close to thespeed of light in a laboratory frame, it has become custom to quote the speed of the particles in terms of their energy--particle physicists routinely talk about "1 GeV" electrons to refer to the kinetic + rest mass energy of the electron in question, rather than bothering to translate this into a translational velocity.
${}^{1}$ This is true unless $m=0$. In the latter case, you can get a finite answer out of the $0\cdot \infty$ in the case of massless particles, but ONLY if they travel at exactly at the speed of light. Were they to travel faster than this, the formula would no longer be the indeterminate form $0 \cdot \infty$, but rather $0\cdot ({\rm complex \,\,number})$, which would be a definite, but nonsensical answer.