Both expressions are valid, because definitions are not correct or wrong. Definitions are consensus that we take about elements in a formalism.
We can define anything as we want in a formalism, simply we have to respect basic logic rules, such as using always the same definition, once we chose one. Moreover, definitions must have physical meaning in theories of physical content.
So we have to ask ourselves, what is the physical meaning of kinetic energy? Usually, we define kinetic energy as the energy that changes due to work and both definitions of kinetic energy
$$E_{kin}^{(1)} \equiv mc^2 (\gamma -1)$$
$$E_{kin}^{(2)} \equiv mc^2 \gamma$$
are physically acceptable according to this criteria, because their time derivatives equal the rate of doing work
$$\frac{dE_{kin}^{(1)}}{dt} = \frac{dE_{kin}^{(2)}}{dt} = \mathbf{F}\mathbf{v}$$
which is expected because the difference between both definitions is a constant term, which vanish when differentiating.
In the momentum representation
$$E_{kin}^{(1)} \equiv \sqrt{m^2c^4 + \mathbf{p}^2c^2} - mc^2$$
$$E_{kin}^{(2)} \equiv \sqrt{m^2c^4 + \mathbf{p}^2c^2} $$
Most books use the first definition, but I have seen books using the definition $E_{kin}^{(2)}$. A reason for choosing this definition is given by Richard Talman in Geometric Mechanics; Toward a Unification of Classical Physics (Wiley 2007):
Here, we have used the symbol $E_{kin}$ which, since it includes the rest energy, differs by that much from being a generalization of the "kinetic energy" of Newtonian mechanics. Nevertheless it is convenient to have a symbol for the energy of a particle that accompanies its very existence and includes its energy of motion but does not include any "potential energy" due to its position in a field of force.
An example of article using the second definition of relativistic kinetic energy is Decay constants in the heavy quark limit in models à la Bakamjian and Thomas:
also for the kinetic energy $K(\{\mathbf{k}_i\})$, that can be taken to be of the non-relativistic $\frac{\mathbf{k}_i^2}{2m_i}$ or relativistic $\sqrt{\mathbf{k}_i^2 + m_i^2}$ forms.