I seek an inductive proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}).$ I am stuck.
This question is from [Number Theory George E. Andrews 1-1 #3].     Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$   This problem is driving me crazy. $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$  $(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out    This is a similar problem in the book and I tried this method but it wasn't working out      $\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$    $\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$     $\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$.  $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$ \eqalign{ \sum^{(k+1)-1}_ {j=0}x^j &amp; = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\ &amp;= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\ &amp;= \dfrac{x^{k+1}-1}{x-1}. }$$   Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.    $\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m&gt;1$, then $n&lt;m^n.$
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