Prove that 7ⁿ-1 is divisible by 7 for all positive integers n.

Substitute $n=1$ into the expression:

$$7^{1-1} = 7^0 = 1$$

For 1 to be divisible by 7, 1 must be equal to $7k$ for some integer $k$. However, $1/7$ is not an integer.
Therefore, 1 is not divisible by 7.
This means the statement "$7^{n-1}$ is divisible by 7" is false for $n=1$.

For any integer $n \ge 2$, the exponent $n-1$ will be a positive integer.
Let $m = n-1$. Since $n \ge 2$, we have $m \ge 1$.
The expression becomes $7^m$.
We can rewrite $7^m$ as:

$$7^m = 7 \times 7^{m-1}$$

Since $m \ge 1$, the exponent $m-1$ is a non-negative integer ($m-1 \ge 0$).
Therefore, $7^{m-1}$ is an integer.
Let $k = 7^{m-1}$. Since $7^{m-1}$ is an integer, $k$ is an integer.
So, we have:

$$7^m = 7k$$

This shows that $7^m$ is a multiple of 7, and thus divisible by 7.
Substituting back $m = n-1$, we conclude that $7^{n-1}$ is divisible by 7 for all integers $n \ge 2$.

Conclusion:
The statement "$7^{n-1}$ is divisible by 7" is not true for all positive integers $n$. It is false for $n=1$. However, it is true for all integers $n \ge 2$.