Proving that $8^n-2^n$ is a multiple of $6$ for all $n\geq 0$ by induction I have the following induction problem: 

$8^n-2^n$ is a multiple of $6$ for all integers $n\geq 0$.

So far this is what I've done:
Base case: $n = 0$
$8^0-2^0 = 6$
$1 - 1 = 6$
$0 = 6$ 
This means that it's a multiple of $6$.
Assume $n = k\colon 8^k - 2^k = 6m$
Where I'm getting stuck now is factoring out $8^k - 2^k = 6m$. The example I'm following on YouTube was $6^k + 4 = 5m$, which factored out nicely to $(5m-4) \cdot 6 + 4$. Where do I go from here?
 A: For $n\geq 0$, let $S(n)$ denote the statement
$$
S(n) : 6 \mid (8^n-2^n)\Longleftrightarrow 8^n-2^n=6m, m\in\mathbb{Z}.
$$
Base case ($n=0$): $S(0)$ says that $6\mid (8^0-2^0)$, and this is true.
Inductive step: Fix some $k\geq 0$ and assume that $S(k)$ is true where
$$
S(k) : 6\mid (8^k-2^k)\Longleftrightarrow 8^k-2^k=6\ell, \ell\in\mathbb{Z}.
$$
To be shown is that $S(k+1)$ follows where
$$
S(k+1) : 6\mid (8^{k+1}-2^{k+1})\Longleftrightarrow 8^{k+1}-2^{k+1}=6\eta, \eta\in\mathbb{Z}.
$$
Beginning with the left-hand side of $S(k+1)$,
\begin{align}
8^{k+1}-2^{k+1} &= 8\cdot 8^k-2\cdot 2^k\tag{by definition}\\[0.5em]
&= 8(8^k-2^k)+6\cdot 2^k\tag{rearrange}\\[0.5em]
&= 8(6\ell)+6\cdot 2^k\tag{by $S(k)$, the ind. hyp.}\\[0.5em]
&= 6(8\ell+2^k)\tag{factor out $6$}\\[0.5em]
&= 6\eta,\tag{$\eta=8\ell+2^k; \eta\in\mathbb{Z}$}
\end{align}
we end up at the right-hand side of $S(k+1)$, completing the inductive step.
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 0$. $\blacksquare$
A: First of all, your base case should be $0=6\times0$, thus it is a multiple of $6$.
Now as @Brenton said, our inductive step is as follows:
Assume $6$ divides $8^k-2^k$, for some $k$ (we write this $6\mid8^k-2^k$).
We want to show that $6\mid8^{k+1}-2^{k+1}$, using this assumption.
Ok so $$6\mid8^k-2^k$$ so $$6\mid8(8^k-2^k)=8^{k+1}-8\times2^k=8^{k+1}-4\times2^{k+1}$$
Can you finish the proof from here?
A: By the binomial theorem, $8^n=(6+2)^n=6a+2^n$.
Induction is used to prove the binomial theorem.
