Determine all polynomials $P(x)$ with degree $n\geq 1$ and integer coefficients so that for every real number $x$ the following condition is satisfied 
$$P(x)=(x-P(0))(x-P(1))(x-P(2))\cdots (x-P(n-1))$$
1. **Base Case: \( n = 1 \)**

   For \( n = 1 \), the polynomial \( P(x) \) must satisfy:
   \[
   P(x) = (x - P(0))
   \]
   Since \( P(x) \) is a polynomial of degree 1, we can write \( P(x) = ax + b \). Given the condition, we have:
   \[
   ax + b = x - P(0)
   \]
   Comparing coefficients, we get \( a = 1 \) and \( b = -P(0) \). Therefore, \( P(x) = x \) and \( P(0) = 0 \). Hence, \( P(x) = x \) is the only solution for \( n = 1 \).

2. **Inductive Step: \( n > 1 \)**

   **Claim:** If \( n > 1 \), then \( P(0) = 0 \).

   **Proof:** Assume \( P(0) \neq 0 \). Plugging \( x = 0 \) into the given condition, we get:
   \[
   P(0) = (0 - P(0))(0 - P(1))(0 - P(2)) \cdots (0 - P(n-1))
   \]
   Simplifying, we have:
   \[
   P(0) = (-P(0))(-P(1))(-P(2)) \cdots (-P(n-1)) = (-1)^n P(0) P(1) P(2) \cdots P(n-1)
   \]
   Dividing both sides by \( P(0) \) (since \( P(0) \neq 0 \)), we get:
   \[
   1 = (-1)^n P(1) P(2) \cdots P(n-1)
   \]
   This implies that \( P(k) = \pm 1 \) for \( k = 1, 2, \ldots, n-1 \). However, since \( P(1) \neq 0 \), we must have \( P(k) = -1 \) for all \( k = 1, 2, \ldots, n-1 \). This would imply:
   \[
   P(x) = x(x + 1)^{n-1}
   \]
   However, this polynomial does not satisfy the given condition for \( n \ge 2 \). Therefore, \( P(0) \) must be 0.

3. **Further Analysis for \( n \ge 3 \)**

   **Claim:** For \( k = 3, 4, \ldots, n-1 \), any potential solution would have \( P(k) = 0 \).

   **Proof:** Plugging \( x = k \) into the given condition, we get:
   \[
   P(k) = k(k - P(0))(k - P(1))(k - P(2)) \cdots (k - P(n-1))
   \]
   Since \( P(0) = 0 \), this simplifies to:
   \[
   P(k) = k \cdot (k - P(1)) \cdot (k - P(2)) \cdots (k - P(n-1))
   \]
   Let \( T_k = \prod_{\substack{1 \le i \le n-1 \\ i \neq k}} (k - P(i)) \). Then:
   \[
   P(k) = k \cdot (k - P(k)) \cdot T_k
   \]
   Rearranging, we get:
   \[
   (1 + kT_k) P(k) = k^2 T_k
   \]
   Since \( 1 + kT_k \) shares no factors with \( k^2 T_k \), this can only occur if \( 1 + kT_k \in \{-1, 1\} \) or \( T_k = 0 \). For \( k \ge 3 \), we extract \( P(k) = 0 \).

4. **Final Form of \( P(x) \)**

   Given \( P(0) = 0 \) and \( P(k) = 0 \) for \( k = 3, 4, \ldots, n-1 \), we have:
   \[
   P(x) = x(x - P(1))(x - P(2)) x^{n-3}
   \]
   Plugging in \( x = 1 \) and \( x = 2 \), we get:
   \[
   P(1) = (1 - P(1))(1 - P(2))
   \]
   \[
   P(2) = 2(2 - P(1))(2 - P(2)) \cdot 2^{n-3}
   \]
   Solving these equations, we find that \( P(1) = \frac{2(2^n - 2)}{2^n - 8} \) and \( P(2) = \frac{3 \cdot 2^n}{2^n + 4} \). This implies \( n \neq 3 \), and for \( n > 3 \), \( 2^n + 4 \nmid 3 \cdot 2^n \) (since \( 2^n + 4 \nmid -12 \)). Therefore, there are no other solutions.

The final answer is \( \boxed{ P(x) = x } \).