If $2^n$ balls are divided into piles, they can always be brought into a single pile by a finite number of operations 
$64$ balls are separated into several piles. At each step, one takes two different piles $A$ and $B$, having $p$ and $q$ balls respectively. Suppose $p\ge q$. Then one takes $q$ balls from pile $A$ and puts them in pile $B$. Prove that it is possible to bring all the balls in a single pile. Source-102 Combinatorial problems-Titu Andreescu, Zuming Feng

The book shows that it is possible to do the above thing for any $2^n$ balls using induction. While that is a good solution, I was wondering if my solution is correct:
PROOF:
We induct on the number of piles $n$. This is trivial for $n=1,2$. 
Now assume that it is true for $n=k$, with $k\le 63$. 
For $n=k+1$, we simply take the pile with the lowest number of balls and place them in another pile. Now there are $n=k$ piles of balls and by the inductive hypothesis, these balls can be placed  into a single pile.This ends the proof.
[Q.E.D]
Is my proof above correct?
 A: Your proof is incorrect. Even though it is a nice idea, it misses the subtlety of the problem, and has one inference problem. The inference problem is this:
You write that inducting on the number of piles $n$ is trivial for $n=1,2$. Then you assume that the proposition is true for $n=k$ with $k \leq 63$. You cannot assume that it is true for all $k \leq 63$: the proposition being trivially true for $n=2$ certainly does not imply it is true for, for example, $n=62$. You can, thus far, only assume it is true for $k \in \{1,2\}$.
Furthermore, the real problem with the proof comes here. You wrote:

For $n=k+1$, we simply take the pile with the lowest number of balls and place them in another pile. Now there are $n=k$ piles of balls and by the inductive hypothesis, these balls can be placed  into a single pile.This ends the proof.

But you cannot simply take the pile with the lowest number of balls and place them in another pile. Let's re-read the problem:
You are given piles $A,B$ with $p$ and $q$ balls respectively, where $p \geq q$. This means that $A$ has $p$ balls and $B$ has $q$ balls. The piles are pre-labelled essentially according to which one has more balls.
Then one takes $q$ balls from pile $A$ and puts them in pile $B$. So afterwards, pile $A$ has $p-q$ balls, and pile $B$ has $2q$ balls. 
Note that since $p\geq q$, pile $A$ then has $p-q \geq 0$ balls.
So you cannot move all the balls from the smaller pile to the bigger pile (that would be too easy).
The correct method to proceed is indeed by induction, but it's not as straight-forward as you made it out to be. Even the $n=2$ case, I think, is somewhat tricky. I suggest actually working it out with e.g. two piles of coins to see what happens.
A: Use induction on $n$, where the total number of balls is $2^n$.
For $n=0$, there is only one pile, and there is nothing to show.
Assume the problem can be solved when there are $2^n$ balls, and suppose there are $2^{n+1}$ balls in piles.
Now use induction on the number of piles containing an odd number of balls. (Call these “odd piles.”)
If the are no odd piles, then every pile has an even number of balls, and you can glue pairs of balls together within each pile. You now have $2^n$ glue-balls in piles and can solve the problem by the inductive hypothesis using glue-balls instead of balls. (Every move of balls will move an even number of balls, so no glue-balls will have to be broken up.)
If not, and there is a pile with an odd number of balls, there must be at least two piles with an odd number of balls, because there is an even number of balls in all.
Choose two odd piles and move balls between them. This moves an odd number of balls, and after the move, there are two fewer odd piles. By induction, the resulting problem can be solved.
A: Added: As pointed out in the comments, this is completely wrong, since once you move $q$ balls from a pile with $p$ balls to one with $q$ balls, the piles contain $p-q$ and $2q$ balls, so the next move won’t be $q$ balls again. I’ve posted another answer, which I hope is correct.

(Wrong answer)
Consider two piles with $p\ge q$ balls, respectively. Make as many successive moves from the larger pile to the smaller pile as allowed. Then you end up with two piles (one of which may be empty). One of the two piles has fewer than $q$ balls. (It has $p\,(\mbox{mod }q)$ balls, to be exact). The other has some positive multiple of $q$ balls, so is the larger pile now.
Now start moving balls between the resulting piles in the opposite direction. with the two resulting piles, or if one pile vanished because $q\mid p$, with the one resulting pile and another pile, if there is one. At each round, you reduce the size of the smaller of some two piles, and eventually one of the two piles is reduced to size zero. After some number of moves, there will be only one remaining pile.
The assumption that there are $64$ balls appears to be unneeded.
