If $a_i >0$ ($i=1, 2, \cdots , n$), then $$\left (\frac{a_1}{a_2} \right )^k + \left (\frac{a_2}{a_3} \right )^k + \cdots + \left (\frac{a_n}{a_1} \right )^k \geq \frac{a_1}{a_2}+\frac{a_2}{a_3}+\cdots + \frac{a_n}{a_1}$$ for all $k\in \mathbb{N}$
1. Let \( \frac{a_i}{a_{i+1}} = x_i \) for \( i = 1, 2, \ldots, n \). Note that \( a_{n+1} = a_1 \). This implies that \( x_1 x_2 \cdots x_n = \frac{a_1}{a_2} \cdot \frac{a_2}{a_3} \cdots \frac{a_n}{a_1} = 1 \).

2. Consider the function \( f(x) = x^k \). Since \( k \in \mathbb{N} \), \( f(x) \) is a convex function for \( x > 0 \).

3. By Jensen's Inequality for a convex function \( f \), we have:
   \[
   f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) \leq \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}
   \]
   Applying this to our function \( f(x) = x^k \), we get:
   \[
   \left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right)^k \leq \frac{x_1^k + x_2^k + \cdots + x_n^k}{n}
   \]

4. Multiplying both sides by \( n \), we obtain:
   \[
   n \left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right)^k \leq x_1^k + x_2^k + \cdots + x_n^k
   \]

5. Since \( x_1 x_2 \cdots x_n = 1 \), by the Arithmetic Mean-Geometric Mean (AM-GM) Inequality, we have:
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n} = \sqrt[n]{1} = 1
   \]

6. Therefore, we have:
   \[
   \left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right)^{k-1} \geq 1
   \]

7. Combining the results from steps 4 and 6, we get:
   \[
   x_1^k + x_2^k + \cdots + x_n^k \geq n \left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right)^k \geq x_1 + x_2 + \cdots + x_n
   \]

8. Substituting back \( x_i = \frac{a_i}{a_{i+1}} \), we have:
   \[
   \left(\frac{a_1}{a_2}\right)^k + \left(\frac{a_2}{a_3}\right)^k + \cdots + \left(\frac{a_n}{a_1}\right)^k \geq \frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_n}{a_1}
   \]

Thus, the inequality holds for all \( k \in \mathbb{N} \).

The final answer is \( \boxed{ \left(\frac{a_1}{a_2}\right)^k + \left(\frac{a_2}{a_3}\right)^k + \cdots + \left(\frac{a_n}{a_1}\right)^k \geq \frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_n}{a_1} } \)