Find all functions $f:\mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $n$, there exists an unique positive integer $k$, satisfying $f^k(n)\leq n+k+1$. 
1. **Claim: \( f(n) > n + 1 \)**

   **Proof:**
   Suppose \( f(n) \le n + 1 \) for some \( n \). Then let \( m \) be the \( k \) for \( f(n) \). We have:
   \[
   f^{m+1}(n) \le f(n) + m + 1 \le n + 1 + m + 1 = n + m + 2
   \]
   So, \( k = m + 1 \) is valid for \( n \). However, since \( f(n) \le n + 1 \), we have:
   \[
   f(n) \le n + 1 < n + 2
   \]
   Thus, \( k = 1 \) is also valid, which is a contradiction because \( k \) is supposed to be unique. Therefore, our assumption that \( f(n) \le n + 1 \) must be false, implying:
   \[
   f(n) > n + 1
   \]
   \(\blacksquare\)

2. **Uniqueness of \( k \):**

   If \( k > 1 \) for some \( n \) is valid, then:
   \[
   f^{k-1}(n) < f^k(n) - 1 = n + k
   \]
   This implies that \( k-1 \) is also valid for \( n \), which contradicts the uniqueness of \( k \). Therefore, \( k = 1 \) must be valid for all \( n \).

3. **Bounding \( f(n) \):**

   Since \( k = 1 \) is valid for all \( n \), we have:
   \[
   f(n) \le n + 2
   \]
   Combining this with the first claim \( f(n) > n + 1 \), we get:
   \[
   n + 1 < f(n) \le n + 2
   \]
   The only integer \( f(n) \) that satisfies this inequality is:
   \[
   f(n) = n + 2
   \]

4. **Verification:**

   We need to verify that \( f(n) = n + 2 \) satisfies the original condition. For \( f(n) = n + 2 \), we have:
   \[
   f^k(n) = n + 2k
   \]
   We need to check if there exists a unique \( k \) such that:
   \[
   n + 2k \le n + k + 1
   \]
   Simplifying, we get:
   \[
   2k \le k + 1 \implies k \le 1
   \]
   Since \( k \) must be a positive integer, \( k = 1 \) is the only solution. Thus, \( f(n) = n + 2 \) satisfies the condition.

The final answer is \( \boxed{ f(n) = n + 2 } \)