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?

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, $x-y \mid x^n-y^n$.
