1. **Stating the problem:** We want to analyze the expression $7^n - 1$ where $n \in \mathbb{Z}^+$ (positive integers) and determine its divisibility by 6. 2. **Formula and rules:** To check if $7^n - 1$ is divisible by 6, we need to verify if it is divisible by both 2 and 3, since 6 = 2 \times 3. 3. **Divisibility by 2:** Since 7 is odd, $7^n$ is odd for any positive integer $n$. Therefore, $7^n - 1$ is odd minus 1, which is even. Hence, $7^n - 1$ is divisible by 2. 4. **Divisibility by 3:** We check $7^n - 1 \pmod{3}$. Note that $7 \equiv 1 \pmod{3}$, so: $$7^n \equiv 1^n \equiv 1 \pmod{3}$$ Therefore, $$7^n - 1 \equiv 1 - 1 \equiv 0 \pmod{3}$$ which means $7^n - 1$ is divisible by 3. 5. **Conclusion:** Since $7^n - 1$ is divisible by both 2 and 3, it is divisible by 6 for all positive integers $n$. **Final answer:** $$6 \mid (7^n - 1) \quad \text{for all} \quad n \in \mathbb{Z}^+$$