Is my proof of $x^n-y^n = \ldots$ correct?

Question

I am solving problems from Spivak's calculus book's 1st chapter. Basically Spivak wants me to prove this:

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

Question 1: Is my proof correct?

Question 2: How to make it perfect? Any advise?

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My proof:

I guess Spivak wants to say: $$ x^n-y^n = (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) $$

Let: $$ f(x,y,n) = \left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) $$

Then we can say (is this sub-proof perfect?): $$\begin{split} x^{n+1}-y^{n+1} &= (x-y)\left(\sum_{i=1}^{i=n+1} x^{(n+1)-i}y^{i-1}\right)\\ &= (x-y)\left( \begin{split} &x^{(n+1)-(n+1)} y^{(n+1)-1}\\ &+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1} \end{split} \right)\\ &= (x-y)\left( x^{0} y^{n} + \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1} \right)\\ &= (x-y)\left( y^{n} + \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1} \right)\\ &= (x-y)\left( y^{n} + x\sum_{i=1}^{i=n} x^{n-i}y^{i-1} \right)\\ &= (x-y)\left( y^{n} + xf(x,y,n) \right)\\ \end{split}$$

We can also rewrite what Spivak wants into: $$ x^n-y^n = (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big) $$

We've already proven in [spivak_calc_probs.1.1.2]: $$\begin{split} x^2-y^2 &= (x-y)(x+y)\\ &= (x-y)\left(\sum_{i=1}^{i=2} x^{2-i}y^{i-1}\right)\\ &= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\ \end{split}$$

Then, by induction, we prove that: $$\begin{split} x^2-y^2 &= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\ x^3-y^3 &= (x-y)\Big(y^{3-1} + xf(x,y,3-1)\Big)\\ \vdots\\ x^n-y^n &= (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)\\ \end{split}$$

And since $(x-y)(y^{n-1} + xf(x,y,n-1))$ is only a rewrite of what Spivak wants, i.e. $(x-y)(x^{n-1}$ $+$ $x^{n-2}y$ $+$ $\ldots$ $+$ $xy^{n-2})$, therefore Q.E.D already.

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alternative proof:

Using the distributive axiom: $$\begin{split} & (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(x^{n-n}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(x^{0}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x\left(x^{n-1}y^{1-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x\left(x^{n-1}y^{0} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x\left(x^{n-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=2}^{i=n} x^{n-i+1}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-(i+1)+1}y^{(i+1)-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i-1+1}y^{i+1-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i} - y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\ \end{split}$$

Then using additive associative axiom: $$\begin{split} & x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i} - y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\ &= x^{n} - y^{n} + \left(\sum_{i=1}^{i=n-1} x^{n-i}y^{i} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x^{n} - y^{n} + \left(0\right)\\ &= x^{n} - y^{n} \end{split}$$

Answer 1

A proof by induction (although in this case it's not the simplest approach, as pointed out in comments) would proceed as follows:

1) Base case - show the the expansion is correct for $n=1$. In this case the sum $\sum_{i=1}^{i=n} x^{n-i}y^{i-1}$ has only one term which is $x^0y^0=1$ so this is straightforward.

2) Assume the expansion is correct for a specifc value of $n$, say $n=k$.

3) Find an expression that relates the $n=k+1$ case to the $n=k$ case. You could try:

$\quad x^{k+1}-y^{k+1} = x(x^k-y^k) + (x-y)y^k$

4) Use (2) to expand your expression from (3):

$\quad x^{k+1}-y^{k+1} = x(x-y)\sum_{i=1}^{i=k} x^{k-i}y^{i-1} + (x-y)y^k\\ \quad=(x-y)\left( x\sum_{i=1}^{i=k} x^{k-i}y^{i-1} + y^k\right) \\\quad =(x-y)\left(\sum_{i=1}^{i=k+1} x^{k+1-i}y^{i-1}\right)$

So now you have shown that if the expansion is true for $n=k$ then it is also true for $n=k+1$. Together with your base case $n=1$, this proves by induction that the expansion is true for all $n \in \mathbb{N}$.

Answer 2

More simply, in your first proof, for brevity let $S(n)$ be the sentence $(x-y)f(x,y,n)=x^n-y^n.$ Then, using your calculation that $f(x,y,n+1)=y^n+xf(x,y,n),$ we have $$S(n)\implies (x-y)f(x,y,n+1)=$$ $$=(x-y)(y^n+xf(x,y,n))=$$ $$=xy^n-y^{n+1}+(x)(( x-y)f(x,y,n))=$$ $$=xy^n-y^{n+1}+(x)(x^n-y^n)=$$ $$=x^{n+1}-y^{n+1}.$$ That is, we have $$S(n)\implies S(n+1).$$ And $S(1)$ is true because $f(x,y,1)=1.$ So by induction on $n$ we have $S(n)$ for all $n\in \Bbb N.$

Answer 3

$(a - 1)(a^{n-1} + a^{n-2} + ... + 1) =$
$a^n + a^{n-1} + ... + a - (a^{n-1} + a^{n-2} + ... + 1) =$
$a^n - 1$

Set a = x/y and multiply by y$^n$.