Let  $x = \left( 1 + \frac{1}{n}\right)^n$ and $y = \left( 1 + \frac{1}{n}\right)^{n+1}$ where $n \in \mathbb{N}$. Which one of the numbers  $x^y$, $y^x$ is bigger ?
1. **Define the expressions for \( x \) and \( y \):**
   \[
   x = \left( 1 + \frac{1}{n} \right)^n \quad \text{and} \quad y = \left( 1 + \frac{1}{n} \right)^{n+1}
   \]

2. **Express \( y \) in terms of \( x \):**
   \[
   y = \left( 1 + \frac{1}{n} \right)^{n+1} = \left( 1 + \frac{1}{n} \right) \cdot \left( 1 + \frac{1}{n} \right)^n = \left( 1 + \frac{1}{n} \right) \cdot x
   \]
   Therefore,
   \[
   y = \left( 1 + \frac{1}{n} \right) x
   \]

3. **Calculate \( x^y \):**
   \[
   x^y = x^{\left( 1 + \frac{1}{n} \right) x} = x^x \cdot x^{\frac{x}{n}}
   \]
   Since \( x = \left( 1 + \frac{1}{n} \right)^n \), we can rewrite \( x^{\frac{x}{n}} \) as:
   \[
   x^{\frac{x}{n}} = \left( \left( 1 + \frac{1}{n} \right)^n \right)^{\frac{x}{n}} = \left( 1 + \frac{1}{n} \right)^x
   \]
   Therefore,
   \[
   x^y = x^x \cdot \left( 1 + \frac{1}{n} \right)^x
   \]

4. **Calculate \( y^x \):**
   \[
   y^x = \left( \left( 1 + \frac{1}{n} \right) x \right)^x = \left( 1 + \frac{1}{n} \right)^x \cdot x^x
   \]

5. **Compare \( x^y \) and \( y^x \):**
   From the calculations above, we have:
   \[
   x^y = x^x \cdot \left( 1 + \frac{1}{n} \right)^x
   \]
   and
   \[
   y^x = \left( 1 + \frac{1}{n} \right)^x \cdot x^x
   \]
   Since multiplication is commutative, we can see that:
   \[
   x^y = y^x
   \]

Hence, we conclude that \( x^y = y^x \).

The final answer is \( \boxed{ x^y = y^x } \)