Design of butterworth filter
Now, we use the fact that $-1=e^{j\pi(2k-1)}$ and $j=e^{\frac{j\pi}{2}}$. I have no idea what's he saying here.


Well, he's simply stating mathematical truths:


$f(x)=e^{jx}$ is periodic with $2\pi$, and $e^{j\pi(0-1)}=e^{-j\pi}=-1$. If that's not clear, draw a unit cycle in the complex plane and remember Euler's formula.
$j=e^{\frac{j\pi}{2}}$: same business. $\frac\pi2$ is a quarter cycle, and thus you end up at the top point of the unit circle.


You really need to be familiar with the properties of the complex exponential function, or else you will have a very hard time understanding pole/zero diagrams in general.


  And does he take a normalized cut-off frequency? 


Yes, he does (very explicit on slide 9). 


  If yes, how do i change it to arbitrary cut-off frequency.


Um, same slide, explicitly:


  
    We can design filters for any other cut-off frequency by substituting $s$ by $\frac s{\omega_c} $.