Compressing exponentials

[RobertReid]

My lecture does this, and I'm unsure if it is true, or otherwise I don't understand

 

[Dr.Peterson]

Do you see that the first step is to factor out [MATH]e^\frac{-j2\pi kn}{N}[/MATH]?

Then they found that the second term in the other factor simplifies to [MATH]e^\frac{-j2\pi Nn}{2N} = e^{-j\pi n} = \left(e^{-j\pi}\right)^n = (-1)^n[/MATH].