Answer
Prove that $p(1)$ is true.
Prove that $p(k)$ true implies $p(k+1)$ true.
Work Step by Step
Let's note by $p(n)$ the property we want to prove:
$$p(n):\text{ } 7^n-1=6m.$$
$\textbf{Step 1:}$ Prove $p(1)$ is true.
Set $n=1$ on each side and check if the property is verified:
$$\begin{align*}
&7^1-1=6\\
&6\text{ is a multiple of }6\checkmark\end{align*}$$
Therefore $p(1)$ is true.
$\textbf{Step 2:}$ Prove that if $p(k)$ is true, then $p(k+1)$ is true.
We assumed $p(k)$ is true:
$$p(k):\text{ } 7^k-1=6m.\tag1$$
We have to prove:
$$p(k+1):\text{ } 7^{k+1}-1=6q.\tag2$$
Rewrite $7^{k+1}-1$ and use $7^k-1=6m$:
$$\begin{align*}
7^{k+1}-1&=7^k\cdot 7-1\\
&=7^k(6+1)-1\\
&=6(7^k)+(7^k-1)\\
&=6(7^k)+6m\\
&=6(7^k+m)\\
&=6q.
\end{align*}$$
We got Eq. $(2)$, therefore if $p(k)$ is true, then $p(k+1)$ is also true.
Both steps being proved, the given formula is true for any natural number $n$.
