Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$ 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 $$
\eqalign{
\sum^{(k+1)-1}_ {j=0}x^j & = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\
&= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\
&= \dfrac{x^{k+1}-1}{x-1}.
}$$
  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 $m>1$, then $n<m^n.$

 A: Your induction hypothesis is that
$$\sum_{j=0}^{k-1}x^jy^{(k-1)-j}=\frac{x^k-y^k}{x-y}\;.$$
Now follow the model:
$$\begin{align*}
\sum_{j=0}^{(k+1)-1}x^jy^{k-j}&\overset{(1)}=\left(\sum_{j=0}^{k-1}x^jy^{k-j}\right)+x^ky^0\\
&\overset{(2)}=y\left(\sum_{j=0}^{k-1}x^jy^{(k-1)-j}\right)+x^k\\
&\overset{(3)}=y\cdot\frac{x^k-y^k}{x-y}+x^k\\
&\overset{(4)}=\frac{x^ky-y^{k+1}+x^{k+1}-x^ky}{x-y}\\
&\overset{(5)}=\frac{x^{k+1}-y^{k+1}}{x-y}\;.
\end{align*}$$
$(1)$ is splitting off the last term of the summation; $(2)$ factors a $y$ out of the remaining summation; $(3)$ uses the induction hypothesis; and $(4)$ and $(5)$ are just algebra.
A: $$\frac{1-(x/y)^n}{1-x/y}=1+x/y+(x/y)^2+...+(x/y)^{n-1}$$
$$\frac{(y^n-x^n)/y^n}{(y-x)/y}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$
$$\frac{y^n-x^n}{(y-x)y^{n-1}}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$
$$y^n-x^n=(y-x)(y^{n-1}+xy^{n-2}+...+x^{n-1})$$
A: Hint: Apply $\textbf{Theorem 1.2}$ to $\displaystyle \frac{x}{y}$.
Edit: It follows from $\textbf{Theorem 1.2}$ that $\displaystyle \Bigl(\frac{x}{y}\Bigr)^n -1=\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$.
Now multiply the equation by $y^n$ to get
$$\displaystyle y^n\Bigl(\Bigl(\frac{x}{y}\Bigr)^n -1)\Bigr)=y^n\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$
Simplifying on the left-hand side and rewritting $y^n$ as $yy^{n-1}$ on the right-hand side we get
$$(x^n -y^n)=yy^{n-1}\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$
Because the product is commutative you can rewrite the right-hand side to get
$$(x^n -y^n)=y\Bigl(\frac{x}{y}-1\Bigr)y^{n-1}\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$
Finally, on the right-hand side, factor in $y$ and $y^{n-1}$ accordingly to get
$$(x^n -y^n)=(x-y)(x^{n-1}+\cdots +xy^{n-2}+y^{n-1})$$
A: Let $u_n$ $=$ $x^n-y^n$. Now note that $u_n$ $=$ $(x+y)u_n$$_-$$_1$ $+$ $xy$ $u_n$$_-$$_2$. Assume that the given expression is true for all numbers from $1$ to some fixed $k(>1)$. Then apply induction to prove the result for $(k+1)$. 
