Prove that $(a+b) (a^2 + b^2) (a^4 + b^4)...(a^{32} + b^{32}) = a^{64} - b^{64}$ if $b = a-1$ Prove that if $b = a-1$, then
$(a + b) (a^2 + b^2 ) (a^4 + b^4 ) ... (a^{32} + b^{32} ) = a^{64} - b^{64}$ .
I saw this in a website and it wrote this:
hint: 
Write down the equality $1 = a+b$ and use the formula $k^2 -n^2 = (k-n) (k + n) $
Decision:
We write the equality $1 = a+b$ and use the formula $k^2 -n^2 = (k-n) (k + n)$ (If you are not familiar with the formulas of reduced multiplication, prove this formula: multiply the expression on the right-hand side.). We write the expression $1*(a+b)(a^2+b^2)(a^4+b^4)…(a^{32}+b^{32})= (a-b)(a+b)(a^2+b^2)(a^4+b^4)…(a^{32}+b^{32})= (a^2-b^2)(a^2+b^2)(a^4+b^4)…(a^{32}+b^{32})= (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^{16}+b^{16})(a^{32}+b^{32})= (a^8-b^8)(a^8+b^8)(a^{16}+b^{16})(a^{32}+b^{32})= (a^{16}-b^{16})(a^{16}+b^{16})(a^{32}+b^{32})= (a^{32}-b^{32})(a^{32}+b^{32})=a^{64}-b^{64}$
I don't know how this formula $k^2 - n^2 = (k-n)(k+n)$ come out?
Also, I don't understand the decision part?

hint: $a+b = \dfrac{a^2-b^2}{a-b}, a^2+b^2= \dfrac{a^4-b^4}{a^2-b^2}, ...$, can you see the factors that cancel each other ? 
