Prove by induction: If $h$ and $k$ are any two distinct integers, then $h^n-k^n$ is divisible by $h-k$. If $h$ and $k$ are any two distinct integers, then $h^n-k^n$ is divisible by $h-k$.
Let's start with the basis. Let $n=1$, then 
$h^1-k^1 = h-k$
Now for the induction, I can't use $k$ because I don't want to be confused. So let $P(r)$ for $h^n-k^n$ and that's  $h^r-k^r$
$h^r-k^r = h-k$
$h^r = h-k +k^r$
So, for $P(r+1)$
$h^{r+1}-k^{r+1}$
$h^r * h^1 - k^r * k^1$
$ (h-k +k^r) * h -k^r *k $
This is the point where I'm not certain if I should distribute the $h $ all over the place...so here it is
$ (h*h-k*h +k^r*h)  -k^r *k $
$ (h*h)+(-k*h) +(k^r*h)  -k^r *k $
$  (h)*(h-k) +  (k^r)*(h-k)$
$(h-k) * (h+k^r)$
 A: For $n=0$: $h-k\mid h^0-k^0$.
Suppose $h-k\mid h^n-k^n$, then
$$
\begin{align}
h^{n+1}-k^{n+1}
&=h\cdot h^n-k\cdot k^n\\
&=(\color{#C00000}{h-k})h^n+k(\color{#C00000}{h^n-k^n})
\end{align}
$$
Since $h-k$ and $h^n-k^n$ are divisible by $h-k$, so is $h^{n+1}-k^{n+1}$.
A: Notice that you have
$$
\begin{align}
 &a^2-b^2 = (a-b)(a+b) \\
 &a^3-b^3 = (a-b)(a^2+ab+b^2) \\
 &a^4-b^4 = (a-b)(a^3+a^2b+ab^2+b^3) \\
etc
\end{align}
$$
In fact for every integer $n \ge 1$ you have
$$
 a^n-b^n = (a-b) \cdot \sum_{k=0}^{n-1}a^kb^{n-1-k}
$$
Of course you can prove this formula by induction, or just develop the expression and order the terms of the sum to check the equality.
With this formula you see immediately the divisibility.
A: Another option is to write $$h^{r+1}-k^{r+1}=hh^{r}-kk^{r}=(h+k)(h^r-k^r)-kh(h^{r-1}-k^{r-1})$$
in a similar fashion to what we do to prove that if $\alpha,\beta$ are the roots of $f[X]\in\Bbb Z[X]$, $\alpha^n+\beta^n\in\Bbb Z$ for every $n\in\Bbb Z$. Indeed, $\alpha+\beta,\alpha\beta\in\Bbb Z$ by Vieta and
$$\alpha^{n+1}+\beta^{n+1}=(\alpha+\beta)(\alpha^n+\beta^n)-\alpha\beta(\alpha^{n-1}+\beta^{n-1})$$
