# How can we prove that x^n - 1 always be a multiple of x - 1 [duplicate]

It wonder with a finite series

$$y = x^0 + x^1 + x^2 + ... x^{n-1}$$

can be formulate into $$\frac{x^n - 1}{x - 1}$$

But I don't understand why $x^n - 1$ could be divide by $x-1$ and always be an integer. What is the relation between $x^a - 1$ and $x - 1$. It seem like a mystery

Are there any proof, if possible a visual proof, that would make it easy to understand this relation?

For a visual proof, try thinking in the other direction.

$(x-1)\cdot (x^0+x^1+x^2+\dots+x^{n-1})$

$= x\cdot (x^0+x^1+x^2+\dots+x^{n-1})-1\cdot (x^0+x^1+x^2+\dots+x^{n-1})$

$= (\color{blue}{x^1}+\color{green}{x^2}+\color{red}{x^3}+\dots+\color{orange}{x^{n-1}}+x^n) - (x^0+\color{blue}{x^1}+\color{green}{x^2}+\dots+\color{orange}{x^{n-1}})$

$=-x^0 + (x^1-x^1)+(x^2-x^2)+(x^3-x^3)+\dots+(x^{n-1}-x^{n-1})+x^n$

$=x^n-1$

Maybe the algebraically most obvious way to see it is as follows:

$$\begin{eqnarray} y & = & x^0 + \color{blue}{x^1 + x^2 + ... + x^{n-1}} \\ x\cdot y & = & \color{blue}{x^1 + x^2 + ... + x^{n-1}} + x^n\end{eqnarray}$$ By subtracting the equations you get $$\Rightarrow xy-y = (x-1)y = x^n-1$$

HINT

Let consider

$$(x-1)(x^{n-1}+x^{n-2}+\ldots+x+1)$$