Prove $7$ divides $13^n- 6^n$ for any positive integer I need to prove $7|13^n-6^n$ for $n$ being any positive integer.
Using induction I have the following:
Base case: 
$n=0$:   $13^0-6^0 = 1-1 = 0,   7|0$
so, generally you could say:
$7|13^k-6^k , n = k \ge 1$
so, prove the $(k+1)$ situation:
$13^{(k+1)}-6^{(k+1)}$
$13 \cdot 13^k-6 \cdot 6^k$
And then I'm stuck....where do I go from here?

First, show that this is true for $n=1$:
$13^{1}-6^{1}=7$
Second, assume that this is true for $n$:
$13^{n}-6^{n}=7k$
Third, prove that this is true for $n+1$:
$13^{n+1}-6^{n+1}=$
$13\cdot13^{n}-6\cdot6^{n}=$
$(7+6)\cdot13^{n}-6\cdot6^{n}=$
$7\cdot13^{n}+6\cdot13^{n}-6\cdot6^{n}=$
$7\cdot13^{n}+6\cdot(\color{red}{13^{n}-6^{n}})=$
$7\cdot13^{n}+6\cdot\color{red}{7k}=$
$7\cdot13^{n}+7\cdot6k=$
$7\cdot(13^{n}+6k)$

Please note that the assumption is used only in the part marked red.
