= Solution
For a <periodic potential>, the Hamiltonian commutes with every lattice translation. The three primitive translations are commuting <unitary operators>, so within each <energy eigenspace> one can choose simultaneous translation eigenstates. Their <eigenvalues> have modulus one and can be written $e^{i\boldsymbol k\cdot\boldsymbol a_i}$. Translation by any lattice vector then gives
$$
\psi_{n\boldsymbol k}(\boldsymbol r+\boldsymbol l)=e^{i\boldsymbol k\cdot\boldsymbol l}\psi_{n\boldsymbol k}(\boldsymbol r),\qquad\boxed{\psi_{n\boldsymbol k}=e^{i\boldsymbol k\cdot\boldsymbol r}u_{n\boldsymbol k}(\boldsymbol r),\quad u(\boldsymbol r+\boldsymbol l)=u(\boldsymbol r).}
$$
This proves Bloch's theorem in its eigenbasis form; arbitrary superpositions inside a degenerate eigenspace need not themselves have a single Bloch wavevector.
The <reciprocal lattice> consists of vectors $\boldsymbol g$ with $\boldsymbol g\cdot\boldsymbol a_i\in2\pi\mathbb Z$. Changing $\boldsymbol k$ to $\boldsymbol k+\boldsymbol g$ and $u$ to $e^{-i\boldsymbol g\cdot\boldsymbol r}u$ gives the same physical wavefunction with a periodic new $u$. The <eigenvalue problem> on one cell has a <discrete set> of levels, giving bands indexed by $n$, and this equivalence yields $E_n(\boldsymbol k+\boldsymbol g)=E_n(\boldsymbol k)$. A reciprocal-lattice <fundamental domain>, conventionally a <Brillouin zone>, therefore labels inequivalent wavevectors.
<Periodic boundary conditions> in physical volume $\mathcal V$ give one allowed wavevector per reciprocal-space volume $(2\pi)^3/\mathcal V$. Including the two spin states, there are $2\mathcal V\,d^3k/(2\pi)^3$ states; \b[per unit physical volume] the count is the printed $2\,d^3k/(2\pi)^3$. Each occupied state carries charge $-e$ and <velocity> $\hbar^{-1}\nabla_kE_n$, so
$$
\boxed{\boldsymbol j=-e\frac2{(2\pi)^3}\int_Bn(\boldsymbol k)\boldsymbol v(\boldsymbol k)\,d^3k.}
$$
For a full band, the integral of $\nabla_kE_n$ is a boundary integral. Opposite faces of a reciprocal fundamental cell have equal energies by periodicity and opposite normals, so their contributions cancel. Hence \b[a full band carries zero <electric current>].
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