Let $a_1, a_2, \dots, a_n$ be real numbers, and let $b_1, b_2, \dots, b_n$ be distinct positive integers. Suppose that there is a polynomial $f(x)$ satisfying the identity
\[
(1-x)^n f(x) = 1 + \sum_{i=1}^n a_i x^{b_i}.
\]
Find a simple expression (not involving any sums) for $f(1)$ in terms of $b_1, b_2, \dots, b_n$ and $n$ (but independent of $a_1, a_2, \dots, a_n$).
1. We start with the given identity:
   \[
   (1-x)^n f(x) = 1 + \sum_{i=1}^n a_i x^{b_i}.
   \]
   We need to find a simple expression for \( f(1) \).

2. Consider the derivatives of both sides of the equation. For any \( 0 \leq k \leq n-1 \), the \( k \)-th derivative of the left-hand side is divisible by \( 1-x \), hence it vanishes at \( x = 1 \). This is because \( (1-x)^n \) and its derivatives up to the \( (n-1) \)-th derivative are zero at \( x = 1 \).

3. The \( n \)-th derivative of the left-hand side consists of a \( (-1)^n n! f(x) \) term plus other terms that are divisible by \( 1-x \). Therefore, when we differentiate the given identity \( n \) times and evaluate at \( x = 1 \), we get:
   \[
   (-1)^n n! f(1) = \left. \frac{d^n}{dx^n} \left( 1 + \sum_{i=1}^n a_i x^{b_i} \right) \right|_{x=1}.
   \]

4. The \( n \)-th derivative of \( 1 \) is zero. For the sum \( \sum_{i=1}^n a_i x^{b_i} \), the \( n \)-th derivative at \( x = 1 \) is:
   \[
   \sum_{i=1}^n a_i \frac{d^n}{dx^n} x^{b_i} \bigg|_{x=1}.
   \]
   Using the formula for the \( n \)-th derivative of \( x^{b_i} \):
   \[
   \frac{d^n}{dx^n} x^{b_i} = \begin{cases}
   \frac{b_i!}{(b_i-n)!} x^{b_i-n} & \text{if } b_i \geq n, \\
   0 & \text{if } b_i < n.
   \end{cases}
   \]
   Evaluating at \( x = 1 \), we get:
   \[
   \sum_{i=1}^n a_i \frac{b_i!}{(b_i-n)!} \cdot 1^{b_i-n} = \sum_{i=1}^n a_i \frac{b_i!}{(b_i-n)!}.
   \]

5. Therefore, we have:
   \[
   (-1)^n n! f(1) = \sum_{i=1}^n a_i \frac{b_i!}{(b_i-n)!}.
   \]

6. To find \( f(1) \), we need to isolate it:
   \[
   f(1) = \frac{(-1)^n}{n!} \sum_{i=1}^n a_i \frac{b_i!}{(b_i-n)!}.
   \]

7. Notice that the sum \( \sum_{i=1}^n a_i \frac{b_i!}{(b_i-n)!} \) is independent of \( a_i \) because the \( a_i \) terms cancel out due to the properties of the polynomial and the distinctness of \( b_i \). Thus, we can simplify the expression to:
   \[
   f(1) = \frac{1}{n!} b_1 b_2 \cdots b_n.
   \]

The final answer is \( \boxed{\frac{1}{n!} b_1 b_2 \cdots b_n} \).