Prove that 9ⁿ - 1 is divisible by 8 for all positive integers n using mathematical induction.

For $n=1$, we have $9^1 - 1 = 9 - 1 = 8$.
Since 8 is divisible by 8, the statement is true for $n=1$.

Assume that the statement is true for some positive integer $k$.
This means $9^k - 1$ is divisible by 8.
So, we can write $9^k - 1 = 8m$ for some integer $m$.
Rearranging this, we get $9^k = 8m + 1$.

We need to show that the statement is true for $n=k+1$.
Consider $9^{k+1} - 1$:

Substitute $9^k = 8m + 1$ from the inductive hypothesis:

Factor out 8:

Since $m$ is an integer, $9m+1$ is also an integer. Therefore, $8(9m+1)$ is divisible by 8.

By the principle of mathematical induction, $9^n - 1$ is divisible by 8 for all positive integers $n$.