Let $f(x)=x^n + a_{n-1}x^{n-1} + ...+ a_0 $ be a polynomial of degree $ n\ge 1 $ with $ n$ (not necessarily distinct) integer roots. Assume that there exist distinct primes $p_0,p_1,..,p_{n-1}$ such that $a_i > 1$ is a power of $p_i$, for all $ i=0,1,..,n-1$. Find all possible values of $ n$.
1. **Identify the polynomial and its roots:**
   Let \( f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \) be a polynomial of degree \( n \geq 1 \) with \( n \) integer roots. Given that \( a_i > 1 \) is a power of \( p_i \) for distinct primes \( p_0, p_1, \ldots, p_{n-1} \), we need to find all possible values of \( n \).

2. **Consider small values of \( n \):**
   - For \( n = 1 \):
     \[
     f(x) = x + a_0
     \]
     Example: \( f(x) = x + 2 \) with root \( x = -2 \). Here, \( a_0 = 2 \) which is a power of the prime \( 2 \).

   - For \( n = 2 \):
     \[
     f(x) = x^2 + a_1 x + a_0
     \]
     Example: \( f(x) = x^2 + 3x + 2 = (x + 1)(x + 2) \) with roots \( x = -1 \) and \( x = -2 \). Here, \( a_1 = 3 \) (power of prime \( 3 \)) and \( a_0 = 2 \) (power of prime \( 2 \)).

   - For \( n = 3 \):
     \[
     f(x) = x^3 + a_2 x^2 + a_1 x + a_0
     \]
     Example: \( f(x) = x^3 + 7x^2 + 11x + 5 = (x + 1)^2(x + 5) \) with roots \( x = -1, -1, -5 \). Here, \( a_2 = 7 \) (power of prime \( 7 \)), \( a_1 = 11 \) (power of prime \( 11 \)), and \( a_0 = 5 \) (power of prime \( 5 \)).

   - For \( n = 4 \):
     \[
     f(x) = x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0
     \]
     Example: \( f(x) = x^4 + 11x^3 + 27x^2 + 25x + 8 = (x + 1)^3(x + 8) \) with roots \( x = -1, -1, -1, -8 \). Here, \( a_3 = 11 \) (power of prime \( 11 \)), \( a_2 = 27 \) (power of prime \( 3 \)), \( a_1 = 25 \) (power of prime \( 5 \)), and \( a_0 = 8 \) (power of prime \( 2 \)).

3. **Consider \( n > 4 \):**
   - Assume \( f(x) = (x + y_1)(x + y_2) \cdots (x + y_n) \) where \( y_i \) are positive integers.
   - Given \( a_0 = p^k = y_1 y_2 \cdots y_n \) for some prime \( p \), all \( y_i \) must be powers of \( p \).
   - If at least two \( y_i \) are divisible by \( p \), then \( a_1 \) is also divisible by \( p \), which contradicts the distinct prime condition for \( a_i \).

4. **Contradiction for \( n > 4 \):**
   - Assume \( y_1 = p^k \) and \( y_2 = y_3 = \cdots = y_n = 1 \).
   - Compute coefficients: \( a_k = \binom{n-1}{k-1} + a_0 \binom{n-1}{k} \).
   - For \( k = 2 \):
     \[
     2a_2 = 2(n-1) + a_0(n-1)(n-2)
     \]
     - This implies \( a_2 \) and \( a_{n-2} \) have a common divisor greater than 1, contradicting the distinct prime condition.

Thus, the only possible values for \( n \) are \( 1, 2, 3, 4 \).

The final answer is \( \boxed{ n = 1, 2, 3, 4 } \).