Let $N = 2^{\left(2^2\right)}$ and $x$ be a real number such that $N^{\left(N^N\right)} = 2^{(2^x)}$. Find $x$.
1. First, we need to determine the value of \( N \). Given \( N = 2^{(2^2)} \), we can simplify this as follows:
   \[
   N = 2^{(2^2)} = 2^4 = 16
   \]

2. Next, we are given the equation \( N^{(N^N)} = 2^{(2^x)} \). Substituting \( N = 16 \) into the equation, we get:
   \[
   16^{(16^{16})} = 2^{(2^x)}
   \]

3. We need to simplify the left-hand side of the equation. First, we calculate \( 16^{16} \):
   \[
   16 = 2^4 \implies 16^{16} = (2^4)^{16} = 2^{(4 \cdot 16)} = 2^{64}
   \]

4. Now, substituting \( 16^{16} = 2^{64} \) back into the original equation, we get:
   \[
   16^{(16^{16})} = 16^{(2^{64})}
   \]

5. We know that \( 16 = 2^4 \), so we can rewrite \( 16^{(2^{64})} \) as:
   \[
   16^{(2^{64})} = (2^4)^{(2^{64})} = 2^{(4 \cdot 2^{64})} = 2^{(2^{64 + 2})} = 2^{(2^{66})}
   \]

6. Therefore, the equation \( 16^{(16^{16})} = 2^{(2^x)} \) simplifies to:
   \[
   2^{(2^{66})} = 2^{(2^x)}
   \]

7. Since the bases are the same, we can equate the exponents:
   \[
   2^{66} = 2^x
   \]

8. This implies:
   \[
   x = 66
   \]

The final answer is \( \boxed{66} \)