$(1−x^n) = (1−x)(1 + x + x^2 +\cdots + x^{n−1})$

Question

Is this expression generally true?

$$(1−x^n) = (1−x)(1 + x + \cdots + x^{n−1})$$

The closest Identity I could find in "Mathematical Handbook of Formulas and Tables", Schaum's Outline is:

$$x^{2n+1} - y^{2n+1} = (x-y)(x^{2n} +x^{2n-1}y + x^{2n+2}y^2 + \cdots +y^{2n})$$

Which I could almost believe is the same thing if I set x=1.

$$1 - y^{2n+1} = (1-y)(1 +y + y^2 + \cdots + y^{2n})$$

let $N=2n+1$

$$1 - y^N = (1-y)(1 +y + y^2 + \cdots + y^{N-1})$$

But doesn't this restrict N to be an odd number? I was just asking because i saw this identity being used in a case where N could be an even or odd number, which made me think its generally true. I'm just not sure how to prove it.

Answer 1

Of course it is always true just multiply

$$(1−x)(1 + x + \cdots + x^{n−1}) =$$

$$= 1 + x + \cdots + x^{n−1}-x(1 + x + \cdots + x^{n−1})=$$

$$=1\color{red}{ + x + \cdots + x^{n−1}- x-x^2 - \cdots - x^{n−1}}-x^n=1-x^n$$

Answer 2

Here are the formulæ I learnt in med-school:

$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dots+xy^{n-2}+y^{n-1}),$$ in other words, it is $x-y$ times the sum of all homogeneous monomials in $x$ and $y$, with total degree $n-1$.

The sum of two powers requires an odd exponent: $$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x^{2n-2}y^2-\dots+x^2y^{2n-2}-xy^{2n-1}+y^{2n}).$$

Answer 3

If you expand the sum of powers of $x$ times $1-x$, you get the sum of powers minus the sum of the same powers plus one. After cancellation, all intermediate powers vanish and $1-x^n$ remains.