Drude-Sommerfeld model for the metallic solid: Why does every state occupy a volume of $\frac{\pi^3}{V}$? Textbook: Introduction to Quantum Mechanics 3rd Ed. (Griffiths and Schroeter)
I am an undergrad studying quantum physics for the first time (on the second trimester). We are on Chapter 5 of the Griffiths. In section 5.3.1, Griffiths introduces the Sommerfield "electron gas" model of the solid. 
After solving the Schrodinger equation and finding the form of the energy, Griffiths claims 

If you imagine a three/dimensional space, with axis $k_x,k_y,k_z$, and planes drawn in at $k_x = \frac{n\pi}{l_x},k_y = \frac{n\pi}{l_y},k_z = \frac{n\pi}{l_z}$, each intersection point represents a distinct (one particle) stationary state. Each block in this grid, and hence also each state, occupies a volume $$\frac{\pi^3}{V}$$

I don't see why "Each block in this grid, and hence also each state, occupies a volume" $$\frac{\pi^3}{V}$$. 
 A: $\pi$ seems to appear out of nowhere in this derivation. In my opinion, this should read $\Delta k_d=2\pi/L_d$ ($L_d$ being the side lengths of a 3D bulk cube), and then multiply over all 3 dimensions in k-space to get to a volume in k-space: $(\Delta k_d)^3=8\pi^3/V$.
The $2\pi$ is the normalization constant of the Fourier transformation (which is chosen dependent on the scientific field, usually).
The $\Delta k$'s represent the distance between the quantum mechanical states of a particle in bulk 3D. The density of states in k-space is the inverse of the distance between the states. Usually, the calculation goes on to expressing it as a function of energy $D(E)$ as described here.
A: The vector $\vec{k}$ uniquely labels all possible states. So for your system, the set of possible states form a 3D grid, each point labelled by a unique $\vec{k}$. 
The distance between each grid point is $\pi/l$ and in each volume element of $\pi^3/V$ there is only one k-point.
An analogous case of 2D is graphically shown below. 

