Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$

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

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

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.

$$\frac{1-(x/y)^n}{1-x/y}=1+x/y+(x/y)^2+...+(x/y)^{n-1}$$ $$\frac{(y^n-x^n)/y^n}{(y-x)/y}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$ $$\frac{y^n-x^n}{(y-x)y^{n-1}}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$ $$y^n-x^n=(y-x)(y^{n-1}+xy^{n-2}+...+x^{n-1})$$