Solve the following system of equations:

$$
\left\{\begin{array}{r}
x_{1} \cdot x_{2} \cdot \ldots \cdot x_{n}=1 \\
x_{1}-x_{2} \cdot x_{3} \cdot \ldots \cdot x_{n}=1 \\
x_{1} \cdot x_{2}-x_{3} \cdot x_{4} \cdot \ldots \cdot x_{n}=1 \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
x_{1} \cdot x_{2} \cdot \ldots \cdot x_{n-1}-x_{n}=1
\end{array}\right.
$$
For every $1 \leq i \leq n$, $t_{i}=x_{1} \cdot x_{2} \cdot \ldots \cdot x_{i} \neq 0$, so we can multiply by these numbers. Multiply equation $\left(1_{i}\right)$ by $t_{i}$ $(i=1,2, \ldots, n-1)$. We get

$$
\begin{gathered}
t_{1}^{2}-1=t_{1} \\
t_{2}^{2}-1=t_{2} \\
\cdot \\
\cdot \\
\cdot \\
t_{n-1}^{2}-1=t_{n-1}
\end{gathered}
$$

Thus, each product is a solution to the equation

$$
t^{2}-1=t
$$

so it is equal to one of the numbers

$$
t^{\prime}=\frac{1+\sqrt{5}}{2}, \quad t^{\prime \prime}=\frac{1-\sqrt{5}}{2}
$$

Now, the numbers $t_{1}, t_{2}, \ldots, t_{n-1}$ can be chosen arbitrarily from the numbers $t^{\prime}, t^{\prime \prime}$. The solution to the system is

$$
\begin{aligned}
x_{1} & =t_{1} \\
x_{2} & =\frac{t_{2}}{t_{1}} \\
& \cdot \\
& \cdot \\
& \cdot \\
x_{n-1} & =\frac{t_{n-1}}{t_{n-2}} \\
x_{n}= & \frac{1}{t_{n-1}}
\end{aligned}
$$

These numbers indeed satisfy the system (1), since after substitution, from (1 $\left.1_{0}\right)$ we get an identity, and from $\left(1_{i}\right)$ we get the

$$
t_{i}-\frac{1}{t_{i}}=1
$$

relation, which is a consequence of (2), so (1) is indeed satisfied for the chosen value of $t_{i}$.

The value set of $t_{1}, t_{2}, \ldots, t_{n-1}$ can be chosen in $2^{n-1}$ different ways, so this is the number of solutions to the system. For example, for $n=2$, rounded to three decimal places,

$$
\begin{array}{l|r|r}
x_{1} & t^{\prime}=1.618, & t^{\prime \prime}=-0.618 \\
x_{2} & 1 / t^{\prime}=0.618, & 1 / t^{\prime \prime}=-1.618
\end{array}
$$

For $n=3$,

$$
\begin{array}{l|l|l|l|l}
x_{1} & t^{\prime}=1.618, & t^{\prime}=1.618, & t^{\prime \prime}=-0.618, & t^{\prime \prime}=-0.618 \\
x_{2} & \frac{t^{\prime}}{t^{\prime}}=1, & \frac{t^{\prime \prime}}{t^{\prime}}=-0.384, & \frac{t^{\prime}}{t^{\prime \prime}}=-2.618, & \frac{t^{\prime \prime}}{t^{\prime \prime}}=1 \\
x_{3} & \frac{1}{t^{\prime}}=0.618 ; & \frac{1}{t^{\prime \prime}}=-1.618 ; & \frac{1}{t^{\prime}}=0.618 ; & \frac{1}{t^{\prime \prime}}=-1.618
\end{array}
$$

In any solution, the value of $x_{1}$ is one of $t^{\prime}$ and $t^{\prime \prime}$, the value of $x_{n}$ is one of the reciprocals, and the value of the intermediate unknowns is one of the quotients that can be formed from $t^{\prime}$ and $t^{\prime \prime}$; one of the values $1, -t^{\prime 2}$, and $-t^{\prime \prime 2}$. Nevertheless, each of the $2^{n-1}$ solutions mentioned is different, since if two selections first differ in the value of $t_{k}$, then the corresponding two solutions differ in $x_{k}$.