Proving if $-1 < x < 1$ then $x^1 + x^2 + \cdots + x^n = \frac{x-x^{n+1}}{1 - x}$

Question

Let $$S_n = x + x^2 + x^3 + \cdots + x^n$$ then $$xs_n = x^2 + x^3 + \cdots + x^n + x^{n+1}$$

This is taken from book "An concise introduction to pure mathematics" :

Why does inserting $x$ to left side of equation produce $x^{n+1}$ ?

Answer 1

Simply multiplying all the terms by $x^1$ increases the exponents of $x$ by $1$ in each term. So $x^n\cdot x^1=x^{n+1}.$

Answer 2

$x=1$ is a root of $1-x^{n+1}$ and thus $$1-x^{n+1}=(1-x)(1+x+x^2+\cdots+x^n)\implies 1+x+x^2+\cdots+x^n=\frac{1-x^{n+1}}{1-x}$$$$\implies S_n=\frac{1-x^{n+1}}{1-x}-1=\cdots$$