For $ n \in \mathbb{N}$, let $ f(n)\equal{}1^n\plus{}2^{n\minus{}1}\plus{}3^{n\minus{}2}\plus{}...\plus{}n^1$. Determine the minimum value of: $ \frac{f(n\plus{}1)}{f(n)}.$
1. Define the function \( f(n) \) as follows:
   \[
   f(n) = 1^n + 2^{n-1} + 3^{n-2} + \cdots + n^1
   \]
   We need to determine the minimum value of \( \frac{f(n+1)}{f(n)} \).

2. First, we claim that \( \frac{f(n+1)}{f(n)} > 3 \) for all \( n > 4 \).

3. To prove this, we need to show that:
   \[
   1^{n+1} + 2^n + 3^{n-1} + \cdots + (n+1)^1 > 3(1^n + 2^{n-1} + 3^{n-2} + \cdots + n^1)
   \]

4. Consider the series of inequalities:
   \[
   1^{n+1} + 2^n + 3^{n-1} + \cdots + (n-1)^3 + n^2 + (n+1)^1 > 1^{n+1} + 2^n + 3^{n-1} + (4^{n-2} + 5^{n-3} + \cdots + (n-1)^3 + n^2)
   \]

5. We observe that:
   \[
   (4^{n-2} + 5^{n-3} + \cdots + (n-1)^3 + n^2) > 3(4^{n-3} + 5^{n-4} + \cdots + (n-1)^2 + n)
   \]

6. Therefore, our goal is to prove that:
   \[
   1^{n+1} - 3 \cdot 1^n + 2^n - 3 \cdot 2^{n-1} + 3^{n-1} - 3 \cdot 3^{n-2} + 4^{n-2} - 3 \cdot 4^{n-3} + 5^{n-3} - 3 \cdot 5^{n-4} + \cdots + n^2 - 3 \cdot n > 0
   \]

7. This inequality holds for all \( n > 4 \). Hence, we have:
   \[
   f(n+1) > 3f(n) \quad \forall \, n > 4
   \]

8. Next, we check for \( n = 1, 2, 3, 4 \) to find the minimum value of \( \frac{f(n+1)}{f(n)} \).

9. By inspection:
   - For \( n = 1 \):
     \[
     f(1) = 1, \quad f(2) = 1^2 + 2^1 = 1 + 2 = 3
     \]
     \[
     \frac{f(2)}{f(1)} = \frac{3}{1} = 3
     \]

   - For \( n = 2 \):
     \[
     f(2) = 3, \quad f(3) = 1^3 + 2^2 + 3^1 = 1 + 4 + 3 = 8
     \]
     \[
     \frac{f(3)}{f(2)} = \frac{8}{3}
     \]

   - For \( n = 3 \):
     \[
     f(3) = 8, \quad f(4) = 1^4 + 2^3 + 3^2 + 4^1 = 1 + 8 + 9 + 4 = 22
     \]
     \[
     \frac{f(4)}{f(3)} = \frac{22}{8} = \frac{11}{4} = 2.75
     \]

   - For \( n = 4 \):
     \[
     f(4) = 22, \quad f(5) = 1^5 + 2^4 + 3^3 + 4^2 + 5^1 = 1 + 16 + 27 + 16 + 5 = 65
     \]
     \[
     \frac{f(5)}{f(4)} = \frac{65}{22} \approx 2.95
     \]

10. Clearly, the minimum value of \( \frac{f(n+1)}{f(n)} \) is attained for \( n = 2 \):
    \[
    \left( \frac{f(n+1)}{f(n)} \right)_{\text{min}} = \boxed{\frac{8}{3}}
    \]