statistical mechanics - Why this is the density of points in $k$-space?

I'm reading a solid state physics book and there's something which is confusing me, related to the free electron gas.

After solving Schrodinger's equation with $V = 0$ and with periodic boundary conditions, one finds out that the allowed values of the components of $\mathbf{k}$ are:

$$k_x = \dfrac{2n_x\pi}{L}, \quad k_y=\dfrac{2n_y \pi}{L}, \quad k_z = \dfrac{2n_z\pi}{L}.$$

In the book I'm reading the author says that it follows from this that: there is one allowed wavevector - that is, one distinct triplet of quantum numbers $k_x,k_y,k_z$ - for the volume element $(2\pi/L)^3$ of $\mathbf{k}$ space.

After that he says that this implies that in the sphere of radius $k_F$ the total number of states is

$$2 \dfrac{4\pi k_F^3/3}{(2\pi/L)^3}=\dfrac{V}{3\pi^2}k_F^3 = N,$$

where the factor $2$ comes from spin.

Now, why is that the case? Why it follows from the possible values of $k_x,k_y,k_z$ that density of points in $k$-space? I really can't understand this properly.