Prove the following by using the principle of mathematical induction for all $n \in N:$
$2^{3n}-1$ is divisible by $7$.

(N/A) Let $P(n): 2^{3n}-1$ be divisible by $7$.
Step $1$: Base case for $n=1$:
$P(1) = 2^{3(1)} - 1 = 8 - 1 = 7$,which is divisible by $7$.
So,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in N$:
$P(k) = 2^{3k} - 1 = 7m$ for some integer $m$.
Therefore,$2^{3k} = 7m + 1$ --- $(1)$
Step $3$: Prove $P(k+1)$ is true:
$P(k+1) = 2^{3(k+1)} - 1 = 2^{3k+3} - 1$
$= 2^{3k} \times 2^3 - 1$
$= (7m + 1) \times 8 - 1$ (using equation $(1)$)
$= 56m + 8 - 1$
$= 56m + 7$
$= 7(8m + 1)$
Since $7(8m + 1)$ is a multiple of $7$,$P(k+1)$ is divisible by $7$.
Thus,by the principle of mathematical induction,$2^{3n}-1$ is divisible by $7$ for all $n \in N$.