Non-uniqueness of the k-vector in Bloch state How to understand that Bloch wave solutions can be completely characterized 
by their behaviour in a single Brillouin zone? Given Bloch wave:
\begin{equation*}
\psi_{\mathbf{k}}(\mathbf{r}) = u_{\mathbf{k}}(\mathbf{r}) \exp (i\mathbf{k}\mathbf{r})
\end{equation*}
I can write wavefunction for momentum $\mathbf{k}' = \mathbf{k} + \mathbf{G}$.
\begin{equation*}
\psi_{\mathbf{k}+\mathbf{G}}(\mathbf{r}) = u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G}\mathbf{r})\exp (i\mathbf{k}\mathbf{r})
\end{equation*}
As I understand $u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G}\mathbf{r}) \neq u_{\mathbf{k}}(\mathbf{r})$, so:
\begin{equation*}
  \psi_{\mathbf{k}+\mathbf{G}}(\mathbf{r}) \neq \psi_{\mathbf{k}}(\mathbf{r}).
\end{equation*}
Or I am wrong?
P.S. 
If Bloch functions are periodic in $\mathbf{k}$ space, then why free particle solution is not a Bloch function ($V=0=V(\mathbf{r}) = V(\mathbf{r} + \mathbf{G})$).
 A: $u_k(r)$ has the periodicity of the crystal. Therefore, its Fourier expansion only includes reciprocal lattice vectors:
$$u_k(r)= \sum_GC_{k-G}e^{iG\cdot r}.$$
Therefore,
$$u_{k'}(r)\exp(iGr)=u_k(r).$$
A: Let's consider the simplest problem: electron in free space. As free space is completely periodic, we can take an arbitrary value $a$ as a period of the potential. That is
$$V(x)=\textrm{const.}=V(x+a),$$
which allows the application of Blochs theorem.
Now Schrödinger's equation for such electron looks (in dimensionless units) like
$$-\psi''(x)=K^2\psi(x),\tag1$$
where $K$ is (vacuum) wavenumber of the electron, $K^2=E$ is total energy. Using Bloch's theorem, we can substitute $\psi(x)=u(x)e^{ikx}$:
$$-u''(x)-2iku'(x)+k^2u(x)=K^2u(x).\tag2$$
Here $k$ is quasiwavenumber. General solution of $(2)$ is
$$u(x)=A\exp(-i(k+K)x)+B\exp(-i(k-K)x).\tag3$$
By Bloch's theorem we know that $u(x)=u(x+a)$, the same goes for its derivative. Thus, we can impose periodic boundary conditions on $(2)$ and find the particular solution. From $u(0)=u(a)$ we can get using $(3)$:
$$A=B\left(\frac{e^{2iaK}-1}{e^{ia(k+K)}-1}-1\right),\tag4$$
thus
$$\frac{u(x)}B=e^{-i(k-K)x}+e^{-i(k+K)x}\left(\frac{e^{2iaK}-1}{e^{ia(k+K)}-1}-1\right).\tag5$$
$B$ is normalization constant, for our purposes we can ignore it, taking $B=1$.
Now, from $u'(0)=u'(a)$ we have restriction on $K$:
$$K=k-\frac{2\pi n}a,\,n\in\mathbb{Z}.\tag6$$
Substituting $(6)$ into $(5)$, we get:
$$u(x)=\exp\left(-i\frac{2\pi n}ax\right).$$
Finally, we get our Bloch solution for free electron:
$$\psi(x)=\exp\left(-i\frac{2\pi n}ax\right)\exp(ikx)=\exp\left(i\left(k-\frac{2\pi n}a\right)x\right).$$
You may see that this function is parametrized by two values: $k\in\mathbb{R}$ and $n\in\mathbb{Z}$. $k$ is quasiwavenumber, while $n$ indexes branches of $E(k)$ dependence.
For each $k$ we have a set of possible energies, each corresponding to some band. These bands are uniquely (for given $k$) determined by $n$. Now what happens if you replace $k\to k-G=k-\frac{2\pi}a$? Simple:
$$K_n=k-\frac{2\pi n}a\;\to\; \left(k-\frac{2\pi}a\right)-\frac{2\pi n}a,$$
and right hand side of the above is nothing but
$$RHS=\left(k-\frac{2\pi}a\right)-\left(\frac{2\pi (n+1)}a-\frac{2\pi}a\right)=k-\frac{2\pi(n+1)}a=K_{n+1}.$$
I.e. if you increase $k$ so far as to go out of first Brillouin zone, you just end up in another band. I.e. you don't get new states this way.
This just means that the band structure can be seen in extended zones scheme:

(picture source)
If you look at the picture above, you'll see that the fact that you appeared in a different band is just an artifact of our band indexing and the form of the solution. You could as easily continuously move along the lines in the picture periodically. And it appears that this motion is actually more natural for crystals, due to the places where the bands split. Here's how it looks for non-free electron in crystal:

