Proving nth difference of squares I'm a little stuck on proving $$(b^n-a^n)=(b-a)(a^0b^{n-1}+a^1b^{n-1}+ \cdots + a^{n-1}b^{0}).$$A solution I came across gave this as an answer:
$$(b-a)(b^n + b^{n-1} a+...+ba^{n-1} +a^{n} )\\
=(b-a)b^n + (b-a)b^{n-1} a+...+(b-a)ba^{n-1} +(b-a)a^n\\
=b^n+1 -b^n a+b^n a-b^{n-1} a^2 +...+b^2 a^{n-1} -ba^n +ba^n -a^{n+1}\\
=b^{n+1} -a^{n+1}$$
Since for each $i=0,\ldots,n$, notice that for each term $b^{n-i} a^{i}$ there is a
$-b^{n-i} a^{i}$, so everything cancels except for $b^{n+1} -a^{n+1}$ , so
$$(b-a)(b^n + b^{n-1} a+\cdots +ba^{n-1} +a^n )= b^{n+1} - a^{n+1}$$
But how does proving that $b^{n+1} - a^{n+1}$ is true prove the formula? 
 A: If it's true for $n+1$ then it is true for $m=n+1$.  And if is true for $m $ then it is true for $n=m $.
Or, more generally, if you prove it for any arbitrary value, you have proven it for all arbitrary values.  $n$ (for $n \ge 1$) and $n+ 1$ (for $n \ge 0$) or $k +357$ for ($k \ge -356$) are equally arbitrary.
$(b-a)(b^{k+356}+ b^{k+355}a+...+ba^{k+355}+ a^{k+356})=$
$b^{k+357}-a^{357}) $
Is it not clear we have proven it for all variables?
Wha if we were asked to prove $x-y $ divides $x^h - y^h $?  Would we have to prove it all over again just because we proved it for $a,b,n+1$ but not for $y,x,h $?
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"Unfortunately I had trouble coming up with an alternative proof that gave back $b^n -a^n$"
Seriously???
$(b-a)(a^0b^{n-1} + a^1b^{n-2}+...+a^{n-2}b^1+a^{n-1}b^0)=$
$b ( (a^0b^{n-1} + a^1b^{n-2}+...+a^{n-2}b^1+a^{n-1}b^0) - a (a^0b^{n-1} + a^1b^{n-2}+...+a^{n-2}b^1+a^{n-1}b^0)=$
$b^n +(ab^{n-1}-ab^{n-1}) +....+(a^{n-1}b-a^{n-1}b) - a^n=$
$b^n-a^n $
You weren't able to modify the proof to do it for one lower power?
A: To prove that 
$$\begin{align}
b^n-a^n &= (b-a)(a^0b^{n-1}+a^1b^{n-1} + \cdots + a^{n-1}b^{0})\\
&=(b-a)\sum_{k=0}^{n-1}b^ka^{n-1-k}
\end{align}$$
It suffices to show that $$b^n-1 = (b-1)\sum_{k=0}^{n-1}b^k,$$
as follows:
$$\begin{align}
(b-1)\sum_{k=0}^{n-1}b^k &=b\sum_{k=0}^{n-1}b^k-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=0}^{n-1}b^{k+1}-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=1}^{n}b^{k}-\sum_{k=0}^{n-1}b^k\\
&=b^{n}-1\\
\end{align}$$
