<s>[INST] 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? 
 [/INST] 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:
$$\begin{align}
(b-1)\sum_{k=0}^{n-1}b^k &=b\sum_{k=0}^{n-1}b^k-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=0}^{n-1}b^{k+1}-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=1}^{n}b^{k}-\sum_{k=0}^{n-1}b^k\\
&=b^{n}-1\\
\end{align}$$
</s>