Consider each number being in some box. If $b$ is a box then let $n(b)$ be the number in the box. Now choose a box $b$ such that $n(b)$ is minimal initially. We will show that $b$ can never move to the right. For each box $c$ that is $m$ places to the right of $b$, consider the quantity $2^{m-1}n(c)$ we will show that this quantity is always greater than or equal to $n(b)$, we will do this by induction: base case: clearly true at the start as $n(b)$ is minimal induction step: if it holds at one point, then after any valid move it will hold again. This can be seen by considering the different moves that can happen: (i) $b$ swaps with its left neighbour. (ii) some two neighbours to the right of $b$ swap (iii) some two nieghbours to the left swap. (we can't have b swap with its right neighbour since by hypothesis it contains a not smaller number). Now that we have this we can easily just induct on the length of the row.