Again, start by understanding the dynamics of the problem. What does increasing \(n\) do? Let's try with a few examples. If \(n = 1\text{,}\) then yes, \(6^1 - 1 = 5\) is a multiple of 5. What does incrementing \(n\) to 2 look like? We get \(6^2 - 1 = 35\text{,}\) which again is a multiple of 5. Next, \(n = 3\text{:}\) but instead of just finding \(6^3 - 1\text{,}\) what did the increase in \(n\) do? We will still subtract 1, but now we are multiplying by another 6 first. Viewed another way, we are multiplying a number which is one more than a multiple of 5 by 6 (because \(6^2 - 1\) is a multiple of 5, so \(6^2\) is one more than a multiple of 5). What do numbers which are one more than a multiple of 5 look like? They must have last digit 1 or 6. What happens when you multiply such a number by 6? Depends on the number, but in any case, the last digit of the new number must be a 6. And then if you subtract 1, you get last digit 5, so a multiple of 5.
The point is, every time we multiply by just one more six, we still get a number with last digit 6, so subtracting 1 gives us a multiple of 5. Now the formal proof:
Proof.
Let \(P(n)\) be the statement, “\(6^n - 1\) is a multiple of 5.” We will prove that \(P(n)\) is true for all \(n \in \N\text{.}\)
Base case: \(P(0)\) is true: \(6^0 -1 = 0\) which is a multiple of 5.
Inductive case: Let \(k\) be an arbitrary natural number. Assume, for induction, that \(P(k)\) is true. That is, \(6^k - 1\) is a multiple of \(5\text{.}\) Then \(6^k - 1 = 5j\) for some integer \(j\text{.}\) This means that \(6^k = 5j + 1\text{.}\) Multiply both sides by \(6\text{:}\)
\begin{equation*}
6^{k+1} = 6(5j+1) = 30j + 6\text{.}
\end{equation*}
But we want to know about \(6^{k+1} - 1\text{,}\) so subtract 1 from both sides:
\begin{equation*}
6^{k+1} - 1 = 30j + 5\text{.}
\end{equation*}
Of course \(30j+5 = 5(6j+1)\text{,}\) so is a multiple of 5.
Therefore \(6^{k+1} - 1\) is a multiple of 5, or in other words, \(P(k+1)\) is true. Thus, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \N\text{.}\)