<s>[INST] Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$. Exercise

Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$.


I've posted my solution below. In case someone has a more clever solution, feel free to post it!
(TBH, I was surprised that there was no question on Math.SE regarding this equation!)
 [/INST] $a^n - b^n =$
$= a^{n}b^{0} + (a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1}) \\ \space\space\space - a^{0}b^{n} - (a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1}) = $
$= (a^{n}b^{0} + a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1}) \\ \space\space\space - (a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1} + a^{0}b^{n}) = $
$= a(a^{n-1}b^{0} + a^{n-2}b^{1} + \cdots + a^{1}b^{n-2} + a^{0}b^{n-1}) \\ \space\space\space - b(a^{n-1}b^{0} + a^{n-2}b^{1} + \cdots + a^{1}b^{n-2} + a^{0}b^{n-1}) = $
$= (a - b)(a^{n-1}b^{0} + a^{n-2}b^{1} + \cdots + a^{1}b^{n-2} + a^{0}b^{n-1})= $
$= (a - b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
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