<s>[INST] Use induction to prove that $3 |(7^n −1)$ for every natural number n. This is what I have so far and now I'm getting lost. 
Proof - We prove by induction. Let $P(n)$ be the statement "$3|(7^n - 1)$". Since $3|6$, we see that $P(1)$ holds. Suppose that is true for $n = k$. We must show the result is true for $n = k + 1$. 
Consider,
$$ 7^{(k+1)} - 1 = (7-1)(7^k + 7^{(k-1)} + \cdots + 1)
                 = 6(7^k + 7^{(k-1)} + \cdots + 1)$$
I do not know if I am doing it correctly or if I assumed my goal. 
 [/INST] Your proof is valid if we assume that we know that $a^{m+1} - b^{m+1}=(a^m + ba^{m-1} + ... + b^{m-1}a + b^m)$.
But if we do know that then your proof is one line:  $7^n - 1=(7-1)(7^{m-1}+...+1) = 6*(7^{m-1}+...+1)$ and no need for induction at all.
So I'm assuming you aren't supposed to know that result.  But could you prove the result? 
To do it without the result we can do
Induction step: Assume $6|7^k-1$
then $7^{k+1}-1 = 7*7^k- 1 = (6+1)7^k -  1=$
$6*7^k + (7^k -1)$ and $6|6*7^k$ and $6|7^k-1$ so $6|6*7^k + (7^k -1)$
....
or you could just prove the result
You can prove it by induction:
$a^1-b^1 = (a-b)$
Induction step:
Assume $a^k - b^k = (a-b)(a^{k-1} + a^{k-2}b + ... +ab^{k-2} + b^{k-1})$
Then $a^{k+1} -b^{k+1} =(a*a^{k} - a*b^k)+(a*b^k- b*b^k)=$
$a(a^k-b^k) +(a-b)b^k =$
$(a-b)a(a^{k-1} + a^{k-2}b + ... + ab^{k-2} + b^{k-1}) + (a-b)b^k=$
$(a-b)[a(a^{k-1} + a^{k-2}b + ... + ab^{k-2} + b^{k-1}) + b^k]=$
$(a-b)[(a^k + a^{k-1}b + .... +a^2b^{k-2} + ab^{k-1}) + b^k]=$
$(a-b)(a^k + a^{k-1}b + .... +a^2b^{k-2} + ab^{k-1} + b^k$
Now the result is proven you can use it.
.....
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