I think I understand now. After m moves, the top 2m cards are now either discarded or below the (2^n+m)^th card. Only the even ones are below the (2^n+m)^th card, and the last one of those even cards (also the last card) would be 2m. The last card in a pile of 2^n cards will be the last card remaining since in a stack of cards with a power of 2 number of cards, it will be a skip the first, keep the second, skip the third, keep the fourth, ... skip the (2^n-1)^th, keep the (2^n)^th then repeat this process and only the (2^n)th card will always be kept. 

Thanks!
Tina Jin