Mathematical induction:

Week 1 : Mathematical induction: P(n) => P(n+1) for all natural numbers n

P(n) : ∀ n ∈ ℕ, 7^{n -}1 is a multiple of 6

Method:

First prove base cases.
This is usually P(0) and/or other cases that we cannot prove by assuming the case before it is true. For example, we can’t assume P(-1) is true to prove P(0) since -1 is not a natural number. In this problem, the only base case we need to prove is when n = 0
So, P(0) = 70 -1 = 0 = 6 * 0
Since the result of P(0) confirms the predicate, then the base case is true

Then comes the inductive step where we assume for an arbitrary natural number n, P(n) is true.
Under this assumption, we write down the elements involved in P(n+1) in terms of that of P(n).

             P(n+1) = 7ⁿ⁺¹ -1 = 7( 7ⁿ ) - 1

Since we know that P(n) is true, that means 7ⁿ = 6k + 1 (#7ⁿ -1 is a multiple of 6)
* We write P(n+1) in terms of P(n) because we know that P(n) is already true through the assumption, which would make it easier for us to see how the relationship between P(n+1) and P(n) will hint that P(n+1) is true as well.

Now, we can substitute 7ⁿ as 6k + 1 to see if the result would still be a multiple of 6 in P(n+1)
P(n+1) = 7ⁿ⁺¹ -1 = 7(6k + 1) -1 = 42k + 6 = 6(7k + 1)

This shows that 7ⁿ +1 -1 is also a multiple of 6 and causing P(n+1) to be True.
Since n was considered to be just an arbitrary natural number in the assumption, we can extend the conclusion to say that for any natural number, if P(n) is true, then P(n+1) is true.
Therefore, for any natural number n, 7ⁿ -1 is a multiple of 6