## Exploring the Pattern: (a-1)(a+1)(a2+1)(a4+1)

This expression holds a fascinating pattern that leads to a surprisingly simple result. Let's delve into it:

### Recognizing the Difference of Squares

Notice that the first two factors, (a-1) and (a+1), are in the form of a difference of squares:

**(a-1)(a+1) = a² - 1²**

Similarly, the next two factors can be manipulated to fit the same pattern:

**(a² + 1)(a⁴ + 1) = (a²)² + 1²**

### Applying the Difference of Squares Pattern Repeatedly

Now, we can apply the difference of squares pattern again:

**a² - 1² = (a + 1)(a - 1)****(a²)² + 1² = (a² + 1)(a² - 1)**

Substituting these back into the original expression:

**(a-1)(a+1)(a² + 1)(a⁴ + 1) = (a + 1)(a - 1)(a² + 1)(a² - 1)**

### Simplifying the Expression

Continuing the process of applying the difference of squares pattern, we get:

**(a + 1)(a - 1)(a² + 1)(a² - 1) = (a² - 1²)(a⁴ - 1²) = (a² - 1)(a⁴ - 1)**

Finally:

**(a² - 1)(a⁴ - 1) = a⁶ - 1**

### The Final Result

Therefore, the simplified form of the expression **(a-1)(a+1)(a2+1)(a4+1)** is **a⁶ - 1**. This pattern demonstrates the power of recognizing and applying algebraic identities to simplify complex expressions.