Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd The question:

Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.

Hence, prove $7 | 6^n + 8^n \iff n ~$ is odd
I realise that this is a proof by induction, and this is what I have so far:
\begin{align}
f(n) & = 6^n + 8^n \\
& = (2\cdot 3)^n + (2^3)^n \\
& = 2^n \cdot 3^n + 2^{3n} \\
\end{align}
\begin{align}
\text{Assume} ~ f(k) & = 6^k + 8^k \\
& = 2^k \cdot 3^k + 2^{3k} \\
f(k+1) & = 2^{k+1} \cdot 3^{k+1} + 2^{3(k+1)} \\
\end{align}
Where do I go from here?
 A: This is my first attempt at answering my own question. I realise the question is of a low quality, but I'm focusing on the question-and-answer aspect. 
There are a few issues with the OP's attempt. Firstly, the base step was skipped. Secondly, and crucially, $f(k+1)$ is no longer odd, so it cannot be used in the inductive step.

Problem: Prove that $6^n+8^n$ is divisible by $7$ iff $n$ is odd.
Solution: Let $f(n)$ denote the statement
\begin{align}
f(n) & = 6^n + 8^n \\
\end{align}
Base Step ($n=1$): 
\begin{align}
f(1) & = 6^1+8^1 \\
& = 6+8 \\
& = 14 \\
\end{align}
$7 | 14 \implies f(1)$ holds.
Inductive Step $f(k) \to f(k+2)$: Fix some $k$, where $k$ is odd. Assume that
$$f(k) = 6^k + 8^k$$
holds true. To be proved is that
$$f(k+2) = 6^{k+2} + 8^{k+2}$$
follows. Beginning with the left-hand side of $f(k+2)$,
\begin{align}
6^{k+2} + 8^{k+2} & = 6^k \cdot6^2 + 8^k \cdot 8^2 \\
& = 6^k \cdot 36 + 8^k \cdot 64 \\
& = 6^k + 6^k \cdot 35 + 8^k + 8^k \cdot63 \\
& = 6^k + 8^k + 6^k \cdot 35 + 8^k \cdot63 \\
& = f(k) + 7(6^k\cdot5+8^k \cdot9)
\end{align}
by the inductive assumption, $7 | f(k)$. Clearly, $7 | (6^k\cdot5+8^k \cdot9)$, thereby showing $f(k+2)$ is true, completing the inductive step.
Conclusion: By mathematical induction, it is proved that $6^n+8^n$ is divisible by $7$ iff $n$ is odd. $\Box$
A: $6\equiv1\pmod7, ~8\equiv-1\pmod7$, hence $$6^n+8^n\equiv1^n+(-1)^n\pmod7$$ $1^n+(-1)^n=0~\iff~n$ is odd.
A: Because $$6^n+8^n=(6+8)(6^{n-1}-6^{n-2}\cdot8+...-6\cdot8^{n-2}+8^{n-1}).$$
