Simple proof of $x^n−y^n=(x−y)(x^{n−1}+x^{n−2}y+…+xy^{n−2}+y^{n−1})$

Question

I'd like to see a simple proof of $$x^n−y^n=(x−y)(x^{n−1}+x^{n−2}y+…+xy^{n−2}+y^{n−1})$$

I don't know what proof by induction is... is there a simpler way? The previous post on this doesn't really solve my problem.... I don't understand the induction proofs.

The text that I am reading with this is this:

Answer 1

Note that we can write $$(x−y)(x^{n−1}+x^{n−2}y+…+xy^{n−2}+y^{n−1})=(x-y)\left(\sum_{k=0}^{n-1}x^{n-1-k}y^k\right)$$

We distribute the $(x-y)$ factor over the sum and obtain $$\sum_{k=0}^{n-1}(x^{n-k}y^k-x^{n-1-k}y^{k+1})$$ Now, we will split this up into two sums, and shift the indexing of the second sum. $$\sum_{k=0}^{n-1}x^{n-k}y^k-\sum_{k=0}^{n-1}x^{n-1-k}y^{k+1}=\sum_{k=0}^{n-1}x^{n-k}y^k-\sum_{k=1}^{n}x^{n-k}y^{k}$$ Now, the formulas inside the sums are the sum. If we take off the first term of the first sum, and the last term of the last sum, the remaining terms will cancel: $$\sum_{k=0}^{n-1}x^{n-k}y^k-\sum_{k=1}^{n}x^{n-k}y^{k}=x^n+\sum_{k=1}^{n-1}(x^{n-k}y^k)-\sum_{k=1}^{n}(x^{n-k}y^{k})-y^n=x^n-y^n$$

Answer 2

The simplest proof I know consists in proving first the special case $$1-t^n=(1-t)(1+t+\dots+t^{n-1})$$ by an easy induction:

Suppose the formula is true for some $n$. Then \begin{align} 1-t^{n+1}&=(1-t^n)+(t^n-t^{n+1})=(1-t)(1+t+\dots+t^{n-1})+t^n(1-t)\\ &=(1-t)(1+t+\dots+t^{n-1}+t^n). \end{align}

For the general case, set $y=\dfrac yx$ and rewrite \begin{align} x^n-y^n&=x^n(1-t^n)=x^n(1-t)(1+t+\dots+t^{n-1})\\ &=x(1-t)\,x^{n-1}(1+t+\dots+t^{n-1})\\ &=(x-t)(x^{n-1}+x^{n-2}y+\dotsm+y^{n-1}) \end{align} since $\;x^{n-1}t^i=x^{n-1-i}y^i$ for all $i=0,1,\dots,n-1$.