Is there a way to formulate relativistic dynamics in a way that "hides" the finite speed of light? I'm not referring to the dimensional choice that makes $c=1$; rather I'm imagining something more about replacing all references that apparently involve velocities with the appropriate $\gamma$ factors or rapidity. In this description, the kinematic feature that maps to speed would then be defined on $[0,\infty)$, and we wouldn't have to deal with the question of "why is the speed of light finite, and has the value that it has?" $^*$
$^*$ Or, at least we wouldn't have to deal with it as a fundamental question about the universe.
 A: If you need equations without velocities, just with $\gamma$, just replace any $v$ by $c\sqrt{\gamma^2-1}/\gamma$ and you're done.
Incidentally, $\gamma$ isn't the only useful function of $v$ that takes values between $0$ and $\infty$. You could also use the rapidity $\eta={\rm arctanh}(v/c)$.
It's good to use $\gamma$ and $\eta$ and in various situations, it simplifies physics or makes it clearer. However, it's still true that $d\vec x / dx^0=\vec v$ and nothing else for the trajectory of a particle, so this basic defining equation of the slope of the trajectory takes the simplest form with $\vec v$! And this "actual velocity" is smaller than $c$ for all massive bodies: this basic statement of relativity is surely not just a convention and can't be hand-waved away by some change of variables. Only one quantity, $v$, deserves to be called the velocity.
Also note that $\gamma$ and even $\eta$, if defined in the easy way above isn't even a vector: it only knows about the magnitude of the speed, not its direction.
A: Use a pure four-space notation $x^0 = ct$. Expressions such as $(1/c) (d/dt)$ transform to $(d/dx^0)$. For instance the typical $(v/c)$ terms transform to $(dx/dx^0)$.
For proper-time derivatives use $ds = cd\tau$.
