Problem 1535

1. Let $\omega_k = e^{2 i \pi k/n}$ for $k = 1, 2, \ldots, n-1$. These points are the $n$-th roots of unity excluding $1$.

2. The $n$-th roots of unity are the solutions to the equation $x^n = 1$. These roots can be written as $1, \omega_1, \omega_2, \ldots, \omega_{n-1}$, where $\omega_k = e^{2 i \pi k/n}$.

3. The polynomial whose roots are the $n$-th roots of unity is $x^n - 1$. This can be factored as:
$$x^n - 1 = (x - 1)(x - \omega_1)(x - \omega_2) \cdots (x - \omega_{n-1}).$$

4. We are interested in the product of the distances of the points $\omega_k$ from $1$. This is equivalent to evaluating the product:
$$\prod_{k=1}^{n-1} |1 - \omega_k|.$$

5. To find this product, consider the polynomial $P(x) = x^n - 1$. We can write:
$$P(x) = (x - 1)(x - \omega_1)(x - \omega_2) \cdots (x - \omega_{n-1}).$$

6. Evaluating $P(x)$ at $x = 1$, we get:
$$P(1) = 1^n - 1 = 0.$$

7. However, we need to consider the derivative of $P(x)$ at $x = 1$ to find the product of the distances. The derivative $P'(x)$ is:
$$P'(x) = n x^{n-1}.$$

8. Evaluating the derivative at $x = 1$, we get:
$$P'(1) = n \cdot 1^{n-1} = n.$$

9. The product of the distances from $1$ to each $\omega_k$ is given by the absolute value of the leading coefficient of the polynomial formed by the roots $\omega_k$:
$$\prod_{k=1}^{n-1} |1 - \omega_k| = |P'(1)| = n.$$

Therefore, the product of the distances of these points from $1$ is $n$.

The final answer is $\boxed{n}$