Combinatorics: Sequence of $n$ elements containing only $1$ and $-1$

Question

I have found this problem in an old textbook. We have a sequence of $n$ elements containing only $1$s and $-1$s (eg. $1,1,-1,1,-1,-1$). We make a new sequence by doing the following: first element of the new sequence is obtained by multiplying the first and the second element of the old sequence; second element of the new sequence is obtained by multiplying the second and the third element of the old sequence, and so on. The $n$-th element of the new sequence is obtained by multiplying the last and the first element of the old sequence. With this new sequence we do the same. If $n = 2^k$ prove that eventually we will get a sequence containing only $1$s. In the book it says that this problem should be solved by induction, but I don't see how. The base of induction is not hard to prove, but I am struggling with the inductive step. None of the sub-sequences in $2^{k + 1}$ match the terms of the inductive hypothesis. Can someone please help?

Answer 1

In the $m$-th iteration, the $i$-the entry is equal to $$ \prod_{r=0}^m a_{i+r}^{\binom{m}{r}} $$ where $a_1,...,a_n$ is the sequence in the beginning and indices are taken modulo $n$.

Now we consider what happens after $n=2^k$ iterations. One may see that $\binom{2^k}{r}$ is always even for $1\leq r\leq 2^k-1$. Therefore $a_{i+r}^{\binom{2^k}{r}}=1$ for $1\leq r\leq 2^k-1$ and thus the $i$-the entry will be $$ \prod_{r=0}^{2^k} a_{i+r}^{\binom{2^k}{r}}=a_i\cdot 1\cdots 1\cdot a_{i+2^k}=a_i^2=1. $$