Let us enumerate the balls in the initial positions from $0$ to $2^{k}-1$. Enumerate the positions themselves as the same numbers in that order as well. In particular, every position or ball can be identified with a binary string of length exactly $k$. The little one considers all the pairs of positions $(x, y)$ differing in exactly one bit (in binary). She performs the operation on all these pairs (in any order). Fixing position of the "different bit" we can see that we have $k \cdot 2^k$ such choices. Since each pair corresponds to two choices, we have $k \cdot 2^{k-1}$ such pairs. Thus, it suffices to prove that this is a winning strategy. Now consider a swap performed on a pair (of positions) $(x, y)$ (with $x > y$), which operated on the $i$th bit (i.e. $x-y = x \oplus y = 2^i$ etc). If the number $x'$ now at $x$ has $i$th bit set to zero, then the magician made a move on this turn, otherwise he faked it. To see why this works: note that every pair of (bit, number) is altered by only a unique pair $(x, y)$; thus using the new number we can determine if the bit was changed or not, i.e. the magician faked a move or not.