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?
 A: Note 
$$
13\times 13^k -6 \times 6^k = 7\times 13^k +6(13^k-6^k)
$$
A: There's always this:
$$\begin{split}
x^2-y^2 &= (x-y)(x+y)\\
x^3-y^3 &= (x-y)(x^2+xy+y^2)\\
x^4-y^4 &= (x-y)(x^3 +x^2y+xy^2+y^3)\\
&\vdots\\
x^n-y^n &= (x-y)\left(\sum_{i=1}^{n}x^{n-i}y^{i-1}\right) \mid n\in \mathbb{N} \end{split}$$
In the above, let $x=13$ and $y=6$.  Then you can visually see that no matter what $n\in \mathbb{N}$ is, $ 7\mid 13^n-6^n$.  And more generally, $x-y \mid x^n-y^n$.
A: Hint: $$13\equiv 6\bmod 7$$
rise both sides to the power of $n$.
A: Write $13=6+7$ to expand $13*13^k-6*6^k$.
A: 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.
A: Write $13^n-6^n=(14-1)^n-(7-1)^n$.  Expand using the binomial theorem; the $1$'s cancel and you're left only with multiples of $7$. 
A: We write 
$n=1$ and get
$13 - 6 =7$( a multiple of $7$)
Moving to the induction hypothesis, 
We assume that   $n=k$, therefore 
$13^k - 6^k = 7b$($b$ shows that $7$ is a multiple of $13^k -6^k$ by $b$ times) 
Taking  $k=1$ from above ($n=k=1$) then we get :
$13^1 - 6^1=13 - 6=7$.
Take one side of the equation and use the equivalent to its multiple in each in  the form of $13-7$ or  $7+6$ : $\pm$ the divisor($7$).
Either  $13^k(7+6) - 6^k(6)$ or $13^k(13) - 6^k(13-7)$.
Taking $  13^k(13) - 6^k(13 -7)$ gives
  $13^k(13) - 6^k(13) + 6^k(7)$.
Change the power $k$ on $6^k(13)$ to $13$ ;$13^k(6)$
         $13^k(13) - 13^k(6) + 6^k(7)$
Factor out $13^k$ on the LHS
 $13^k(13-6) + 6^k(7)$ giving
$13^k(7) + 6^k(7)$. This can be compressed to:
$7(13^k + 6^k)$
Therefore now $b= 13^k + 6^k$  making it $7|(13^n - 6^n)$ valid for all positive integers values of $n$.
