Hanging a painting on $n$ nails such that if any one of the nails is removed, the painting falls. Estimate the number of "moves" required.
We show by induction that $\left[\cdots\left[a_{1}, a_{2}\right] \cdots, a_{n}\right]$ is a solution. Our word is therefore

$$
\left[\cdots\left[a_{1}, a_{2}\right], \cdots, a_{n-1}\right] a_{n}\left(\left[\cdots\left[a_{1}, a_{2}\right], \cdots, a_{n-1}\right]\right)^{-1} a_{n}^{-1}
$$

If the nail $n$ is removed, the sequence of words becomes a commutator followed by its inverse, which reduces to 1. If another nail is removed, both commutators reduce to 1, as well as the entire solution, which is what we wanted.

Let $x_{n}$ be the number of moves required. We have $x_{2}=4$ and $x_{n}=2 x_{n-1}+2$ if $n \geqslant 3$. Therefore, $x_{3}=10, x_{4}=22, x_{5}=46$. We then show by induction that $x_{n}=3 \cdot 2^{n-1}-2$. This results in exponential growth, which is not the best.

Instead of separating the last nail, it is better to "separate in the middle." We only consider the case where $n$ is a power of 2: $n=2^{k}$. We then have the word $\left[S, S^{\prime}\right]$, where $S$ is a solution for the first $2^{k-1}$ nails and $S^{\prime}$ is a solution for the last $2^{k-1}$ nails. We then have $x_{2^{k}}=4 x_{2^{k-1}}$ and $x_{2}=4$, from which we deduce $x_{2^{k}}=2^{2 k}$. This is quadratic growth, which is much better than exponential.