Use induction on $k$. Result clear for $k=1$. Assume it is true for some $k > 1$ and now consider an arbitrary $n$-tuple $(x_1, x_2, \dots, x_n)$ of length $n = 2^{k-1}$. Since $x_i^2 = 1$ for all $i$, the second iteration
$$(x_1 x_2^2 x_3, x_2 x_3^2 x_4, \dots, x_{n-1} x_n^2 x_1, x_n x_1^2 x_2)$$
can be written as
$$(x_1 x_3, x_2 x_4, x_3 x_5, \dots, x_{n-1} x_1, x_n x_2)$$
which is the result of the interlacing of the two $(n-1)$-tuples
$$(x_1 x_3, x_3 x_5, \dots, x_{n-1} x_1) \quad \text{and} \quad (x_2 x_4, x_4 x_6, \dots, x_n x_2). \quad (3)$$
The same rule can be used to obtain the fourth iteration of the original $n$-tuple by interlacing the second iteration of the two $(n-1)$-tuples of (3), the sixth iteration of the original by interweaving the third iterations etc.. Thus $2j$ iterations ($j \ge 2$) of the original $n$-tuple yields the same result as the interlacing of the $j^{th}$ iterations of the two $(n-1)$-tuples of (3). But the induction hypothesis guarantees that these iterations of the (3) tuples consist only of ones for sufficiently large $j$. Thus we conclude that, for $j$ sufficiently large, $2j$ iterations of the original $n$-tuple gives the $n$-tuple $(1, 1, \dots, 1)$ as required.