Let $f: \mathbb{R}\to\mathbb{R}$ be a function such that  $(f(x))^{n}$ is a polynomial for every integer $n\geq 2$. Is $f$ also a polynomial?
1. **Given Condition**: We are given that \( f: \mathbb{R} \to \mathbb{R} \) is a function such that \( (f(x))^n \) is a polynomial for every integer \( n \geq 2 \). We need to determine if \( f \) is also a polynomial.

2. **Analyzing \( f^2 \)**: Since \( (f(x))^2 \) is a polynomial, let us denote it by \( P(x) \). Therefore, we have:
   \[
   f(x)^2 = P(x)
   \]
   This implies that \( f(x) = \pm \sqrt{P(x)} \).

3. **Analyzing Higher Powers**: Consider \( f^6 \):
   \[
   f^6 = (f^3)^2 = (f^2)^3 = P(x)^3
   \]
   Since \( f^6 \) is a polynomial, \( P(x)^3 \) must also be a polynomial. This implies that \( P(x) \) itself must be a polynomial.

4. **Continuity of \( f \)**: Since \( f(x) = \pm \sqrt{P(x)} \) and both \( f^3 \) and \( \sqrt[3]{\cdot} \) are continuous, \( f \) must be continuous. 

5. **Behavior Around Zeros of \( P \)**: Consider the zeros of \( P(x) \). Let \( t \) be a zero of \( P(x) \). In the neighborhood of \( t \), \( f(x) \) must be either \( P(x) \) or \( -P(x) \) on intervals where \( P(x) \) does not vanish. 

6. **Consistency Across Intervals**: Suppose \( f(x) = P(x) \) on some interval \( [t, t + \varepsilon) \) and \( f(x) = -P(x) \) on the interval \( (t - \varepsilon, t] \). This would imply a discontinuity at \( t \), which contradicts the continuity of \( f \).

7. **Conclusion**: Since \( f(x) \) must be continuous and consistent across intervals, \( f(x) \) must be either \( P(x) \) or \( -P(x) \) everywhere. Therefore, \( f(x) \) itself must be a polynomial.

\(\blacksquare\)

The final answer is \( \boxed{ f } \) is a polynomial.