elementary number theory - Prove by induction that $8^{n} − 1$ is divisible by $7$

Prove by induction that $8^{n} − 1$ for any positive integer $n$ is divisible by $7$.

Hint: It is easy to represent divisibility by $7$ in the following way: $8^{n} − 1 = 7 \cdot k$ where k is a positive integer.

This question confused me because I think the hint isn't true. If $n = 1$ and $k = 2$ for example, then we end up with $7 = 14$ which is obviously invalid. Does this mean the $n \leq k$ in order for the hint to be true.

Answer

Notice the following steps of mathematical induction,

1. Setting $n=1$, $$8^1-1=7$$ above number is divisible by $7$ for $n=1$




2. assume that $8^n-1$ is divisible by $7$ for $n=k$ then $$8^k-1=7m$$
   or $$8^k=7m+1\tag 1$$
   where, $m$ is some integer




3. setting $n=k+1$, $$8^{k+1}-1$$
   $$=8\cdot 8^k-1$$
   
   setting the value of $8^k$ from (1), $$=8(7m+1)-1$$
   $$=7(8m+1)$$

since, $(8m+1)$ is an integer hence the number $7(8m+1)$ is divisible by $7$ thus $8^n-1$ is divisible by $7$ for $n=k+1$

hence, $8^n-1$ is divisible by $7$ for all integers $n\ge 1$