Find all functions $f:\mathbb{N} \to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: [list][*] $f(2)=2$, [*] $f(mn)=f(m)f(n)$, [*] $f(n+1)>f(n)$. [/list]
We need to find all functions \( f: \mathbb{N} \to \mathbb{N} \) such that for all \( m, n \in \mathbb{N} \):
1. \( f(2) = 2 \)
2. \( f(mn) = f(m)f(n) \)
3. \( f(n+1) > f(n) \)

We will prove by induction that \( f(n) = n \) for all \( n \in \mathbb{N} \).

1. **Base Case:**
   - For \( n = 1 \):
     Since \( f: \mathbb{N} \to \mathbb{N} \), \( f(1) \) must be a natural number. 
     Using the multiplicative property \( f(mn) = f(m)f(n) \), we set \( m = n = 1 \):
     \[
     f(1 \cdot 1) = f(1)f(1) \implies f(1) = f(1)^2
     \]
     This implies \( f(1) = 1 \) because the only natural number \( x \) such that \( x = x^2 \) is 1.

2. **Inductive Step:**
   - Assume \( f(k) = k \) for all \( k \leq n \). We need to show that \( f(n+1) = n+1 \).

   - **Case 1:** \( n+1 \) is a prime number.
     - Since \( n+1 \) is prime, it cannot be factored into smaller natural numbers greater than 1. Therefore, we cannot directly use the multiplicative property. However, we know:
       \[
       f(n) < f(n+1)
       \]
       By the inductive hypothesis, \( f(n) = n \). Thus:
       \[
       n < f(n+1)
       \]
       Since \( f(n+1) \) must be a natural number greater than \( n \) and \( f \) is strictly increasing, the smallest possible value for \( f(n+1) \) is \( n+1 \). Hence:
       \[
       f(n+1) = n+1
       \]

   - **Case 2:** \( n+1 \) is a composite number.
     - Suppose \( n+1 = ab \) where \( a, b > 1 \). By the inductive hypothesis, \( f(a) = a \) and \( f(b) = b \). Using the multiplicative property:
       \[
       f(n+1) = f(ab) = f(a)f(b) = ab = n+1
       \]

Since both cases confirm that \( f(n+1) = n+1 \), by induction, we have \( f(n) = n \) for all \( n \in \mathbb{N} \).

\(\blacksquare\)

The final answer is \( \boxed{ f(n) = n } \) for all \( n \in \mathbb{N} \).