$2^k$ of $\pm1$-s placed around ring turn into all $1$-s in $2^k$ steps. Trying to get a simple solution for the following secondary school problem:

Numbers +1 and -1, total amount $2^k$, are arbitrary placed along a ring. Then between any two of neighbouring numbers their product is placed, while the original numbers are wiped out. This is repeated again and again. Prove 
that in no more then $2^n$ of such operations only 1-s will remain.

Here is my own solution, that looks too complex to me. Maybe someone can suggest smth simpler.
Use $\{a_i\}$ $(0\le i\lt 2^k)$ for the original numbers (ordered clockwise),
and $\{b_i^{(m)}\}$ $(0\le i\lt 2^k)$ the numbers after $m$-th operation, while $b_i^{(0)}=a_i$ $(0\le i\lt 2^k)$
Also $\oplus$ will be used for 'cyclic edition', ie addition by modulo $2^k$ (in other words $a \oplus b$ is the remainder of $a+b$ divided by $2^k$: $a \oplus b = (a+b) \& (2^k-1)$). Therefore case $a_{i \oplus 1}$ is the clockwise neighbour of $a_i$, $a_{i \oplus 2}$ is the second clockwise neighbour of $a_i$, etc. In particular $a_{i \oplus 2^k} = a_i$ (full-ring distance).

Prove (by induction) that 
$$
b_i^{(m)} = \prod_{j=0}^{m} a_{i \oplus j}^\binom{m}{j} \qquad (1)
$$
where $\binom{m}{j}$ are binomial coefficients.
Base For $m=0$, expression $(1)$ becomes (assuming $\binom{0}{0}=1$) $b_i^{(0)} = a_i$, which is true.
Transition If $(1)$ folds for $m$, then
$$
b_i^{({m+1})} = b_i^{(m)}b_{i\oplus 1}^{(m)} = \prod_{j=0}^{m} a_{i \oplus j}^\binom{m}{j} \times \prod_{j=0}^{m}a_{i \oplus j \oplus 1}^\binom{m}{j} = 
$$
$$
\prod_{j=0}^{m} a_{i \oplus j}^\binom{m}{j} \times \prod_{j=1}^{m+1}a_{i \oplus j}^\binom{m}{j-1} = a_i \times (\prod_{j=1}^{m} {a_{i \oplus j}^{\binom{m}{j}+\binom{m}{j-1}}}) \times a_{i \oplus m \oplus 1}
$$
Finally, due to $\binom{m}{j}+\binom{m}{j-1}=\binom{m+1}{j}$ and $\binom{m+1}{0}=\binom{m+1}{m+1}=1$ we get
$$
b_i^{({m+1})} = \prod_{j=0}^{m+1} a_{i \oplus j}^{\binom{m+1}{j}}
$$
to complete that proof of (1).

Substituting $m=2^k$ gives
$$
b_i^{({2^k})} = \prod_{j=0}^{2^k} a_{i \oplus j}^{\binom{2^k}{j}} =
a_i \times ( \prod_{j=1}^{2^k-1} a_{i \oplus j}^{\binom{2^k}{j}} ) \times a_{i \oplus 2^k}
$$
But $a_i = a_{i \oplus 2^k}$, therefore 
$$
b_i^{({2^k})} = a_i^2   \prod_{j=1}^{2^k-1} a_{i \oplus j}^{\binom{2^k}{j}}  
$$
Since all the binomial coefficients, except first and last for power $2^k$ are even (see this thread), all $b_i^{({2^k})}$ are ones, qed.

This is certainly not the solution meant for a secondary school student :)  
 A: It is easy to notice the pattern that you are following Pascal's triangle in the terms which accumulate in the product.
It is fairly easy to notice that for early powers of $2$ the entries in the line are all even apart from the end ones, and you can't control the sign of those - the squares arising from the even numbers will all be $+1$.
It is easy to observe that when you get to the $2^n$th term, the two ends reference the same original number reached in opposite directions (clockwise and counterclockwise), and therefore make a final square giving $+1$.
Then there is the question of proving it - the Pascal's triangle bit is easy. The fact that the entries in the relevant rows are all even may be a little more challenging at high-school level.

You can do the evenness of the binomial coefficients as follows with $p,q$ polynomials:
$$(x^n+2xyp(x,y)+y^n)^2=x^{2n}+4x^2y^2p(x,y)^2+y^{2n}+2\times \text{ cross term multiples of }xy=$$$$=x^{2n}+2xyq(x,y)+y^{2n}$$
Start with $x^2+2xy+y^2$ ie $p(x,y)=1$ and use induction.
A: I think you can also do this by a different induction. After two steps the value at position $i$ is the product of the values at positions $i-1$ and $i+1$ (taking account of wrap-around, the value at $i$ gets squared).
This decouples the circle into two versions of the problem for $n-1$ provided you do two steps at a time instead of one. ie twice what we had for $n-1$.
The base case is easy.
