Prove the following by using the principle of mathematical induction for all $n \in N:$
$3^{2n} - 1$ is divisible by $8$.

Let $P(n)$ be the statement that $3^{2n} - 1$ is divisible by $8$.
Step $1$: For $n = 1$,$3^{2(1)} - 1 = 9 - 1 = 8$,which is divisible by $8$. Thus,$P(1)$ is true.
Step $2$: Assume $P(m)$ is true for some $m \in N$,i.e.,$3^{2m} - 1 = 8k$ for some integer $k$. So,$3^{2m} = 8k + 1$.
Step $3$: We need to show $P(m+1)$ is true,i.e.,$3^{2(m+1)} - 1$ is divisible by $8$.
$3^{2(m+1)} - 1 = 3^{2m} \times 3^2 - 1$
$= (8k + 1) \times 9 - 1$
$= 72k + 9 - 1$
$= 72k + 8$
$= 8(9k + 1)$.
Since $8(9k + 1)$ is divisible by $8$,$P(m+1)$ is true.
By the principle of mathematical induction,$P(n)$ is true for all $n \in N$.