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?
Note 

$$
13\times 13^k -6 \times 6^k = 7\times 13^k +6(13^k-6^k)
$$
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 $n$.