quantum mechanics - Bloch wave function orthonormality?

there is this text book that is giving me a hard time for a while now:

It shows that Bloch wave functions can be written as

$$\Psi_{n\vec{k}}\left(\vec{r}\right) = \frac{1}{\sqrt{V}}e^{i\vec k \vec r}u_{n\vec k}\left(\vec r\right),$$ which is fine to me. It also states that the Bloch factors $u_{n\vec k}\left(\vec r\right)$ may be orthonormalized on the (primitive) unit cell volume $V_{UC}$: $$\frac{1}{V_{UC}}\int_{V_{UC}}d^3r u^*_{n\vec k}\left(\vec r\right) u_{n'\vec k}\left(\vec r\right) = \delta_{n n'}$$ .

However, and here starts my problem, it then concludes that therefore, the Bloch functions $\Psi_{n\vec{k}}\left(\vec{r}\right)$ fulfill

$$\int_V d^3r \Psi^*_{n\vec{k}}\left(\vec{r}\right) \Psi_{n'\vec{k'}}\left(\vec{r}\right)=$$ $$=\frac 1 V \sum_\vec{R} \int_{V_{UC}\left(\vec R\right)}d^3r e^{-i\vec k \left(\vec R + \vec r\right)} u^*_{n\vec k}\left(\vec R + \vec r\right) e^{i\vec {k'} \left(\vec R + \vec r\right)} u_{n'\vec{k'}}\left(\vec R + \vec r\right)=$$ $$=\frac 1 N \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \frac{1}{V_{UC}}\int_{V_{UC}} d^3r u^*_{n\vec{k}}\left(\vec r\right) u_{n'\vec{k'}}\left(\vec r\right)=$$ $$=\delta_{\vec{k'}\vec{k}}\delta_{n'n}$$ with lattice vectors $\vec R$ and crystal volume $V = N V_{UC}$.

But I just don't get the last two lines. I mean, the integral in the second last line actually should read $\int_{V_{UC}} d^3r e^{i\left(\vec{k'}-\vec{k}\right)\vec r}u^*_{n\vec{k}}\left(\vec r\right) u_{n'\vec{k'}}\left(\vec r\right)$, shouldn't it? And, if that is true and I didn't miss something important already, I can't understand how that would yield these two $\delta$s ...

I'm almost sure I missed something, but I just desperately keep fail getting it, so any help would be greatly appreciated!

EDIT

Thank you very much for your reactions! However, it seems I failed to state my problem clear enough, so I figured it might be best to tell you what my approach so far was step by step so may be someone can see where I actually go wrong:

Starting with

$$\int_V d^3r \Psi^*_{n\vec{k}}\left(\vec{r}\right) \Psi_{n'\vec{k'}}\left(\vec{r}\right),$$ I partitioned the integration domain $V=NV_{CU}$, thus getting a sum of integrations over the unit cell volume, yielding $$\sum_{\vec R}\int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec{r} + \vec R \right) \Psi_{n'\vec{k'}}\left(\vec{r} + \vec R\right),$$ which happens to be exactly the second line when exploiting $\Psi_{n\vec{k}}\left(\vec{r}\right) = \frac{1}{\sqrt{V}}e^{i\vec k \vec r}u_{n\vec k}\left(\vec r\right)$. I then proceeded using $\Psi\left( \vec r + \vec R\right)=e^{i\vec k \vec R}\Psi\left(\vec r\right)$. But this yields $$\sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right)$$ which disagrees with the third line in the book where the integral is over the Bloch factors $u_{n\vec k}\left(r\right)$ only.

However even assuming this is just a typo (which I'm not so sure of ...), I would be confronted with the integral $\int_{V_{UC}} d^3r e^{i\left(\vec {k'} - \vec k\right)\vec r} u^*_{n\vec{k}}\left(\vec r\right) u_{n'\vec{k'}}\left(\vec r\right)$ and I can't see how those two $\delta$s would arise from that either.

Thank you all again for your reactions and I hope I know actually stated my problem clearly.

Answer

Let $I \sim \sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R} \int_{V_{UC}} d^3r \Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right)$

The term $\sum_{\vec R} e^{i\left(\vec{k'}-\vec{k}\right)\vec R}$ gives you a $\sim \delta(\vec{k} - \vec{k'})$ term.

Now, you have : $\Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right) \delta(\vec{k} - \vec{k'}) \sim e^{i(\vec k'- \vec k).\vec r} u^*_{n\vec k}\left(\vec r\right) u_{n'\vec k'}\left(\vec r\right)\delta(\vec{k} - \vec{k'})$

Now, with the $\delta(\vec{k} - \vec{k'})$ term, $e^{i(\vec k'- \vec k).\vec r}$ becomes 1.

So you have :

$\Psi^*_{n\vec{k}}\left(\vec r\right) \Psi_{n'\vec{k'}}\left(\vec r\right) \delta(\vec{k} - \vec{k'}) = u^*_{n\vec k}\left(\vec r\right) u_{n'\vec k}\left(\vec r\right)\delta(\vec{k} - \vec{k'})$

where we have replaced $k'$ by $k$, in the indice of $u_{n'\vec k}$.

So, finally, after integration on $r$, we get :

$I \sim \delta_{nn'}\delta(\vec{k} - \vec{k'})$