How does one notice that?
How would you prove the identity a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-2}a + b^{n-1}) ? Well, its easier than it seems.Just compute the multiplication on the right, using the distributive property. (a-b)(a^{n-1} + ba^{n-2} + ... + b^{n-2}a + b^{n-1}) = a(a^{n-1} + ba^{n-2} + ... + b^{n-2}a + b^{n-1}) - b(a^{n-1} + ba^{n-2} + ... + b^{n-2}a + b^{n-1} ) = a^{n} + ba^{n-1} + ... + b^{n-2}a^2 + b^{n-1}a - ba^{n-1} - b^2a^{n-2} - ... - b^{n-1}a - b^{n} = a^n - b^n Of course, the answer by Aman Dubey is also correctAman Dubey's answer to How would you prove the identity a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... + b^{n-2}a + b^{n-1})? The formula of the Geometric Series comes from that fact.