Irska

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 $$(x_1{x}_3,x_3{x}_5,\dots ,x_{n-1}{x}_1)\text{and}(x_2{x}_4,x_4{x}_6,\dots ,x_n{x}_2).(3)$$ 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.