Find all $n$-tuples of reals $x_1,x_2,\ldots,x_n$ satisfying the system:
$$\begin{cases}x_1x_2\cdots x_n=1\\x_1-x_2x_3\cdots x_n=1\\x_1x_2-x_3x_4\cdots x_n=1\\\vdots\\x_1x_2\cdots x_{n-1}-x_n=1\end{cases}$$
Given the system of equations:
\[
\begin{cases}
x_1 x_2 \cdots x_n = 1 \\
x_1 - x_2 x_3 \cdots x_n = 1 \\
x_1 x_2 - x_3 x_4 \cdots x_n = 1 \\
\vdots \\
x_1 x_2 \cdots x_{n-1} - x_n = 1
\end{cases}
\]

1. **First Equation:**
   \[
   x_1 x_2 \cdots x_n = 1
   \]

2. **Second Equation:**
   \[
   x_1 - x_2 x_3 \cdots x_n = 1
   \]
   Multiply both sides by \(x_1\):
   \[
   x_1^2 - x_1 x_2 x_3 \cdots x_n = x_1
   \]
   Using \(x_1 x_2 \cdots x_n = 1\):
   \[
   x_1^2 - 1 = x_1 \implies x_1^2 - x_1 - 1 = 0
   \]
   Solving this quadratic equation:
   \[
   x_1 = \frac{1 \pm \sqrt{5}}{2}
   \]

3. **Third Equation:**
   \[
   x_1 x_2 - x_3 x_4 \cdots x_n = 1
   \]
   Multiply both sides by \(x_1 x_2\):
   \[
   x_1^2 x_2^2 - x_1 x_2 x_3 x_4 \cdots x_n = x_1 x_2
   \]
   Using \(x_1 x_2 \cdots x_n = 1\):
   \[
   x_1^2 x_2^2 - 1 = x_1 x_2 \implies x_1^2 x_2^2 = x_1 x_2 + 1
   \]
   Substitute \(x_1^2 = x_1 + 1\) from the first equation:
   \[
   (x_1 + 1)x_2^2 = x_1 x_2 + 1
   \]
   Simplify:
   \[
   x_1 x_2^2 + x_2^2 = x_1 x_2 + 1 \implies x_2^2 (x_1 + 1) = x_1 x_2 + 1
   \]
   Rearrange:
   \[
   x_1 x_2^2 + x_2^2 = x_1 x_2 + 1 \implies x_1 x_2 (x_2 - 1) + (x_2 - 1)(x_2 + 1) = 0
   \]
   Factor:
   \[
   (x_2 - 1)(x_1 x_2 + x_2 + 1) = 0
   \]
   Since \(x_2 \neq 1\):
   \[
   x_1 x_2 + x_2 + 1 = 0
   \]

4. **General Case:**
   Similarly, for the \(k\)-th equation:
   \[
   x_1 x_2 \cdots x_{k-1} - x_k x_{k+1} \cdots x_n = 1
   \]
   Multiply both sides by \(x_1 x_2 \cdots x_{k-1}\):
   \[
   (x_1 x_2 \cdots x_{k-1})^2 - x_1 x_2 \cdots x_n = x_1 x_2 \cdots x_{k-1}
   \]
   Using \(x_1 x_2 \cdots x_n = 1\):
   \[
   (x_1 x_2 \cdots x_{k-1})^2 - 1 = x_1 x_2 \cdots x_{k-1} \implies (x_1 x_2 \cdots x_{k-1})^2 = x_1 x_2 \cdots x_{k-1} + 1
   \]
   This implies:
   \[
   (x_1 x_2 \cdots x_{k-1})^2 = x_1 x_2 \cdots x_{k-1} + 1
   \]

5. **Conclusion:**
   From the above steps, we see that each \(x_i\) must satisfy a quadratic equation similar to \(x_1\):
   \[
   x_i^2 = x_i + 1
   \]
   Solving this quadratic equation for each \(x_i\):
   \[
   x_i = \frac{1 \pm \sqrt{5}}{2}
   \]
   Therefore, each \(x_i\) must be either \(\frac{1 + \sqrt{5}}{2}\) or \(\frac{1 - \sqrt{5}}{2}\).

The final answer is \(\boxed{\left(\frac{1 \pm \sqrt{5}}{2}, \frac{1 \pm \sqrt{5}}{2}, \ldots, \frac{1 \pm \sqrt{5}}{2}\right)}\).