Prove $x^{rs}-1=(x^s-1)(x^{s(r-1)}+…+x^s+1)$

Prove $x^{rs}-1=(x^s-1)(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$

I can see that this is true for simple results such as $x^6$ and $x^9$

How can I prove that this is true in general.

It telescopes. From $$(x^s-1)(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$$ consider $$x^s \times(x^{s(r-1)}+x^{s(r-2)}+...+x^s+1) - (x^{s(r-1)}+x^{s(r-2)}+...+x^s+1)$$ The first term becomes $$(x^{sr}+x^{s(r-1)}+...+x^{2s}+x^s)$$ which just leaves $$x^{sr}-1$$

This is equivalent to the geometric series formula: $$1+y+y^2+\cdots+y^n={y^{n+1}-1\over y-1}$$ and therefore $$(y-1)(1+y+y^2+\cdots+y^n)=y^{n+1}-1$$. Substituting $y=x^s$ and $n=r-1$ we obtain your formula.

If you require a proof of the geometric series formula, see for instance this question and any of the answers (I recommend mine :P, but they're mainly pretty good).

Hint

Let $x^s=y$. Use Ruffini's rule to divide $(y^r-1)/(y-1)$. Finally undo the previous assignment (that is, let $y=x^s$ back again).

Hint

Set $y=x^s$, then $y=1$ is solution of $$y^r-1=0$$ and thus, you can factorise $y^r-1$ by $(y-1)$ what gives $$y^r-1=(y-1)(y^{r-1}+y^{r-2}+...+y+1).$$

Observe $1+x^s+x^{2s}+\ldots+x^{(r-1)s}$ is the sum of the first $r$ terms of a geometric sequence, and then it is equal to $(1-(x^s)^r)/(1-x^s)$ or $(x^{sr}-1)/(x^s-1)$.

Start with $x^r-1=(x-1)(x^{r-1}+\cdots+x+1)$. Write this identity for $x=y^s$. This gives keeping in mind that $(y^s)^r=y^{rs}$ \begin{align} y^{sr}-1=&(y^s-1)(y^{s(r-1)}+\cdots+y^s+1)\end{align}