Let $k\ge2$ be an integer. The function $f:\mathbb N\to\mathbb N$ is defined by
$$f(n)=n+\left\lfloor\sqrt[k]{n+\sqrt[k]n}\right\rfloor.$$Determine the set of values taken by the function $f$.
Let \( k \ge 2 \) be an integer. The function \( f: \mathbb{N} \to \mathbb{N} \) is defined by 
\[ f(n) = n + \left\lfloor \sqrt[k]{n + \sqrt[k]{n}} \right\rfloor. \]
We need to determine the set of values taken by the function \( f \).

1. Define \( a_m = m^k - m + 1 \), where \( m \) is a positive integer. Note that \( a_m \) is an increasing sequence for \( k \ge 2 \).

2. We first show that \( a_m > (m-1)^k \):
   \[ a_m = m^k - m + 1 > (m-1)^k. \]
   Since \( m^k - m + 1 \) is greater than \( (m-1)^k \), we have:
   \[ a_m + \sqrt[k]{a_m} > a_m + m - 1 = m^k. \]
   Therefore,
   \[ \sqrt[k]{a_m + \sqrt[k]{a_m}} > m. \]

3. Next, we show that \( a_m - 1 < m^k \):
   \[ a_m - 1 = m^k - m < m^k. \]
   Thus,
   \[ a_m - 1 + \sqrt[k]{a_m - 1} < a_m - 1 + m = m^k. \]
   Therefore,
   \[ \sqrt[k]{a_m - 1 + \sqrt[k]{a_m - 1}} < m. \]

4. Combining the results from steps 2 and 3, we have:
   \[ \forall n \in [a_m, a_{m+1} - 1], \quad \sqrt[k]{n + \sqrt[k]{n}} \in (m, m+1). \]
   Hence,
   \[ \left\lfloor \sqrt[k]{n + \sqrt[k]{n}} \right\rfloor = m. \]

5. Therefore, for \( n \in [a_m, a_{m+1} - 1] \):
   \[ f(n) = n + m. \]

6. Consequently,
   \[ f([a_m, a_{m+1} - 1]) = [a_m + m, a_{m+1} - 1 + m] = [m^k + 1, (m+1)^k - 1]. \]

7. Hence, the set of values taken by the function \( f \) is:
   \[ f(\mathbb{N}) = \mathbb{N} \setminus \{ n^k \mid n \in \mathbb{N} \}. \]

The final answer is \( \boxed{ \mathbb{N} \setminus \{ n^k \mid n \in \mathbb{N} \} } \).