For which integers $ n>1$ do there exist natural numbers $ b_1,b_2,\ldots,b_n$ not all equal such that the number $ (b_1\plus{}k)(b_2\plus{}k)\cdots(b_n\plus{}k)$ is a power of an integer for each natural number $ k$? (The exponents may depend on $ k$, but must be greater than $ 1$)
1. **Restate the problem**: We need to determine for which integers \( n > 1 \) there exist natural numbers \( b_1, b_2, \ldots, b_n \) not all equal such that the product \( (b_1 + k)(b_2 + k) \cdots (b_n + k) \) is a power of an integer for each natural number \( k \). The exponents may depend on \( k \), but must be greater than 1.

2. **Case 1: \( n \) is composite**:
    - Suppose \( n \) is composite. Let \( p \) be a prime divisor of \( n \), i.e., \( p \mid n \).
    - We can partition the \( n \) numbers \( b_1, b_2, \ldots, b_n \) into groups of size \( p \). Set the numbers within each group to be equal. For example, let \( b_1 = b_2 = \cdots = b_p \), \( b_{p+1} = b_{p+2} = \cdots = b_{2p} \), and so on.
    - This ensures that the product \( (b_1 + k)(b_2 + k) \cdots (b_n + k) \) is always a \( p \)-th power because each group of \( p \) terms contributes a \( p \)-th power to the product.
    - Therefore, if \( n \) is composite, it is possible to find such \( b_1, b_2, \ldots, b_n \).

3. **Case 2: \( n \) is prime**:
    - Suppose \( n \) is prime. We need to show that it is impossible to find \( b_1, b_2, \ldots, b_n \) such that \( (b_1 + k)(b_2 + k) \cdots (b_n + k) \) is a perfect power for all \( k \).
    - Assume for contradiction that such \( b_1, b_2, \ldots, b_n \) exist. Let \( a_1, a_2, \ldots, a_n \) be distinct natural numbers and consider the product \( P(k) = (a_1 + k)^{e_1} (a_2 + k)^{e_2} \cdots (a_n + k)^{e_n} \), where \( e_1, e_2, \ldots, e_n \) are positive integers summing to \( n \).
    - We want to show that \( P(k) \) cannot be a perfect power for all \( k \).
    - Choose large primes \( p_1, p_2, \ldots, p_n \) such that \( p_i \gg \max |a_i - a_j| \) for all \( i \neq j \).
    - By the Chinese Remainder Theorem, select \( k \) such that \( k \equiv p_i - a_i \pmod{p_i^2} \) for each \( i \).
    - This choice ensures that \( p_i \nmid a_j + k \) for any \( j \neq i \), and thus \( \nu_{p_i}(P(k)) = e_i \).
    - Since \( \gcd(e_1, e_2, \ldots, e_n) = 1 \) (it must divide \( n \) but cannot equal \( n \) because \( n \) is prime), \( P(k) \) cannot be a perfect power.
    - Therefore, if \( n \) is prime, it is impossible to find such \( b_1, b_2, \ldots, b_n \).

Conclusion:
- If \( n \) is composite, it is possible to find \( b_1, b_2, \ldots, b_n \) such that \( (b_1 + k)(b_2 + k) \cdots (b_n + k) \) is a power of an integer for each \( k \).
- If \( n \) is prime, it is impossible to find such \( b_1, b_2, \ldots, b_n \).

The final answer is \( \boxed{ n } \) is composite.