Prove the statement by the Principle of Mathematical Induction: $3^{2n} - 1$ is divisible by $8$ for all natural numbers $n$.

(N/A) Let $P(n): 3^{2n} - 1$ is divisible by $8$.
Step $1$: For $n = 1$,$P(1) = 3^{2(1)} - 1 = 9 - 1 = 8$,which is divisible by $8$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some natural number $k$,i.e.,$3^{2k} - 1 = 8m$ for some integer $m$. This implies $3^{2k} = 8m + 1$ $(i)$.
Step $3$: For $n = k + 1$,we need to show $P(k + 1)$ is true,i.e.,$3^{2(k+1)} - 1$ is divisible by $8$.
$3^{2(k+1)} - 1 = 3^{2k+2} - 1 = 3^{2k} \cdot 3^2 - 1 = 9 \cdot 3^{2k} - 1$.
Substituting $(i)$,we get $9(8m + 1) - 1 = 72m + 9 - 1 = 72m + 8 = 8(9m + 1)$.
Since $8(9m + 1)$ is divisible by $8$,$P(k + 1)$ is true.
Hence,by the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.