Number & Algebra: Proof by Mathematical Induction

Base Case $n = 1$: $7^1 - 1 = 6 = 6(1)$, which is divisible by $6$. The base step holds.

Induction Hypothesis: Assume the statement is true for $n = k$, where $k \in \mathbb{Z}^+$. That is, $7^k - 1 = 6M$ for some integer $M \in \mathbb{Z}$ (so $7^k = 6M + 1$).

Inductive Step: For $n = k + 1$, consider $7^{k+1} - 1 = 7 \cdot 7^k - 1$.

Substitute hypothesis: $7(6M + 1) - 1 = 42M + 7 - 1 = 42M + 6 = 6(7M + 1)$. Since $M$ is an integer, $7M + 1$ is an integer, so $7^{k+1} - 1$ is divisible by $6$.

Conclusion: Since the statement is true for $n = 1$ and whenever it is true for $n = k$ it is also true for $n = k + 1$, by the principle of mathematical induction $7^n - 1$ is divisible by $6$ for all $n \in \mathbb{Z}^+$.