Prove the statement by the Principle of Mathematical Induction: $2^{3n} - 1$ is divisible by $7$ for all natural numbers $n$.

(N/A) Let $P(n): 2^{3n} - 1$ be divisible by $7$ for all $n \in \mathbb{N}$.
Step $1$: For $n = 1$,$P(1): 2^{3(1)} - 1 = 8 - 1 = 7$,which is divisible by $7$.
Therefore,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$.
That is,$2^{3k} - 1 = 7m$ for some $m \in \mathbb{N}$,which implies $2^{3k} = 7m + 1$.
Step $3$: We need to show $P(k+1)$ is true.
$P(k+1): 2^{3(k+1)} - 1 = 2^{3k} \cdot 2^3 - 1$.
Substituting $2^{3k} = 7m + 1$:
$= (7m + 1) \cdot 8 - 1$
$= 56m + 8 - 1$
$= 56m + 7$
$= 7(8m + 1)$,which is clearly divisible by $7$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.