Let $k$ be a fixed positive integer. The $n$th derivative of $\tfrac{1}{x^k-1}$ has the form $\tfrac{P_n(x)}{(x^k-1)^{n+1}}$, where $P_n(x)$ is a polynomial. Find $P_n(1)$.
1. We start with the function \( \frac{1}{x^k - 1} \). We need to find the \( n \)-th derivative of this function and express it in the form \( \frac{P_n(x)}{(x^k - 1)^{n+1}} \), where \( P_n(x) \) is a polynomial. Specifically, we are asked to find \( P_n(1) \).

2. To proceed, we rewrite \( \frac{1}{x^k - 1} \) as \( \left( \frac{1}{x - 1} \right) f(x) \), where \( f(x) = \frac{x - 1}{x^k - 1} \). Notice that \( f(x) \) is smooth in a neighborhood of \( x = 1 \) and \( f(1) = \lim_{x \to 1} \frac{x - 1}{x^k - 1} = \frac{1}{k} \).

3. Using the product rule repeatedly, we find the \( n \)-th derivative of \( \frac{1}{x^k - 1} \). The dominant term in the \( n \)-th derivative is given by:
   \[
   \left( \frac{(-1)^n n!}{(x - 1)^{n+1}} \right) f(x)
   \]
   plus other terms that have smaller powers of \( (x - 1) \) in their denominators.

4. We can rewrite the \( n \)-th derivative as:
   \[
   \frac{(-1)^n n! f(x) + (x - 1) h(x)}{(x - 1)^{n+1}}
   \]
   where \( h(x) \) is also smooth in a neighborhood of \( x = 1 \).

5. Since \( \frac{1}{x^k - 1} = \left( \frac{1}{x - 1} \right) f(x) \), we can express this as:
   \[
   \frac{1}{x - 1} = \frac{1}{(x^k - 1) f(x)}
   \]
   Therefore, the \( n \)-th derivative can be written as:
   \[
   \frac{(-1)^n n! f(x) + (x - 1) h(x)}{[(x^k - 1) f(x)]^{n+1}}
   \]

6. Simplifying, we get:
   \[
   \frac{(-1)^n n! f(x) + (x - 1) h(x)}{(x^k - 1)^{n+1} f(x)^{n+1}}
   \]

7. Evaluating this expression at \( x = 1 \), we note that \( f(1) = \frac{1}{k} \). Thus, the expression simplifies to:
   \[
   \frac{(-1)^n n! \cdot \frac{1}{k} + 0}{(x^k - 1)^{n+1} \left( \frac{1}{k} \right)^{n+1}} = \frac{(-1)^n n! \cdot \frac{1}{k}}{(x^k - 1)^{n+1} \cdot \frac{1}{k^{n+1}}} = \frac{(-1)^n n! k^n}{(x^k - 1)^{n+1}}
   \]

8. Therefore, \( P_n(1) = (-1)^n n! k^n \).

The final answer is \( \boxed{ (-1)^n n! k^n } \).