Find all function $ f: \mathbb{N}\rightarrow\mathbb{N}$ satisfy $ f(mn)\plus{}f(m\plus{}n)\equal{}f(m)f(n)\plus{}1$ for all natural number $ n$
1. **Initial Assumption and Transformation:**
   We start with the functional equation:
   \[
   f(mn) + f(m+n) = f(m)f(n) + 1
   \]
   We propose a transformation \( f(n) = g(n) + 1 \). Substituting \( f(n) = g(n) + 1 \) into the original equation, we get:
   \[
   g(mn) + 1 + g(m+n) + 1 = (g(m) + 1)(g(n) + 1) + 1
   \]
   Simplifying, we obtain:
   \[
   g(mn) + g(m+n) + 2 = g(m)g(n) + g(m) + g(n) + 2
   \]
   Canceling out the common terms, we get:
   \[
   g(mn) + g(m+n) = g(m)g(n) + g(m) + g(n)
   \]
   Let this be our new functional equation \( F(m, n) \).

2. **Base Case Analysis:**
   We analyze the base cases to find \( g(1) \):
   \[
   F(1, 1): g(2) = g(1)^2 + g(1)
   \]
   \[
   F(1, 2): g(3) = g(1)g(2) + g(1) + g(2)
   \]
   \[
   F(1, 3): g(4) = g(1)g(3) + g(1)
   \]
   \[
   F(2, 2): 2g(4) = g(2)^2 + 2g(2)
   \]

3. **Substitution and Simplification:**
   From \( F(1, 1) \):
   \[
   g(2) = g(1)^2 + g(1)
   \]
   From \( F(1, 2) \):
   \[
   g(3) = g(1)(g(1)^2 + g(1)) + g(1) + (g(1)^2 + g(1)) = g(1)^3 + 2g(1)^2 + 2g(1)
   \]
   From \( F(2, 2) \):
   \[
   2g(4) = (g(1)^2 + g(1))^2 + 2(g(1)^2 + g(1))
   \]
   Simplifying the right-hand side:
   \[
   2g(4) = g(1)^4 + 2g(1)^3 + 3g(1)^2 + 2g(1)
   \]
   From \( F(1, 3) \):
   \[
   g(4) = g(1)(g(1)^3 + 2g(1)^2 + 2g(1)) + g(1) = g(1)^4 + 2g(1)^3 + 2g(1)^2 + g(1)
   \]
   Equating the two expressions for \( g(4) \):
   \[
   2(g(1)^4 + 2g(1)^3 + 2g(1)^2 + g(1)) = g(1)^4 + 2g(1)^3 + 3g(1)^2 + 2g(1)
   \]
   Simplifying, we get:
   \[
   g(1)^4 + 4g(1)^3 + 4g(1)^2 + 2g(1) = g(1)^4 + 2g(1)^3 + 3g(1)^2 + 2g(1)
   \]
   \[
   2g(1)^3 + g(1)^2 = 0
   \]
   Factoring out \( g(1)^2 \):
   \[
   g(1)^2(2g(1) + 1) = 0
   \]
   Thus, \( g(1) = 0 \) or \( g(1) = -\frac{1}{2} \). Since \( g(n) \) maps natural numbers to natural numbers, \( g(1) = -\frac{1}{2} \) is not valid. Therefore, \( g(1) = 0 \).

4. **Induction and Conclusion:**
   If \( g(1) = 0 \), then:
   \[
   g(2) = 0, \quad g(3) = 0, \quad g(n) = 0 \quad \forall n \in \mathbb{N}
   \]
   Hence, \( f(n) = g(n) + 1 = 1 \).

   If \( g(1) = 1 \), then:
   \[
   g(n+1) = g(n) + 1
   \]
   By induction, \( g(n) = n \). Thus, \( f(n) = g(n) + 1 = n + 1 \).

The final answer is \( \boxed{ f(n) = 1 } \) or \( f(n) = n + 1 \).