Here, Greg says the answer is A in the solutions video - which was my initial answer. But later on I figured that it could also be 4^2n therefore k could be less than 8 (4) and greater than 8 (16). which is why I marked D.
but Greg marked my initial deduction i.e. A
Confused now!
Yeah, even I got it wrong initially along with Greg 
Now to the question,
lets assume n=0, then 2^0 =1 and k^0 = 1 regardless of k. So k can be anything. Hence, D is the answer.
Another way to look at it is 2^4= 16
So, 16^n = k^n. If n is even then we cannot know if k=-16 or +16. Again answer is D.
Yes, you’re right. I am deep into the video and someone figured this in that too. Thanks a lot!
