Help with exponents please

this is the last part of a proof i was doing.
2^(3n)*e^(i2πn) + 2^(3n)*e^(i4πn)
= 2^(3n) + 2^(3n)
= 2*2^(3n)
=2^(3n+1)

Can you please explain how they got from the first step to the 2nd step (i'm guessing you let n=0 am i right?)
Also how did they get from the 2nd to last step to the last step, is there some rule of exponents that i never knew about?

So e^(i*2*pi) is a case of Euler's expansion which is e^(i*x)=cos(x)+i*sin(x) so if x=2*pi we have
cos(2*pi)+i*sin(2*pi)=1.
So if you rewrite the equation as 2^(3n) * (e^(i2pi))^n + 2^(3n) * (e^(i2pi))^2n then we have
2^(3n) * (1)^n + 2^(3n) * (1)^2n = 2^(3n) + 2^(3n) since 1 to any power equals 1.
Next, 2^(3n) + 2^(3n) can be grouped together to get 2*2^(3n) which it sounds like you understand. Now looking at this written differently might help 2*2^(3n) = 2^(1) * 2^(3n) and when we multiply two integers with exponents we add the exponents so, 2*2^(3n) = 2^(1) * 2^(3n) = 2^(3n+1).
Hope this helps.

That doesn't make sense, unless you left something out. If you let n = 0, then 2^3n = 2^0 = 1 so your answer would be 2. I think you left something out or wrote something down wrong.

yes the "ln" both sides

when you ln (lawn) both sides the "e^i4πn and e^i2πn disappears.