Answer
$f(n)$ is divisible by $7$ for all values of $n$.
Work Step by Step
Suppose that $f(n)=8^n-1$; we need to prove that $f(n)$ is divisible by $7$.
1. Our aim is to find that $7|8^n-1$ is true for $n=1$
$8^1-1=7m \implies m=1 $
So, it is true for $n=1$.
2. Our aim is to find that $f(n)$ is true for $n=k$.Thus, it will also true for $n=k+1$
$8^{(k+1)}-1=(7)8^k+(8^k-1) \implies 8^{k+1}-1=8^k(8)-1$
This yields: $ 8^{k+1}-1=8^{k+1}-1$
So, it is true for $n=k+1$.
Therefore, $f(n)$ is divisible by $7$ for all values of $n$.
