$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$ I was reading a solution in which author used this identity without giving any hint that how it comes ...{maybe it is obvious}
$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$
by using expansion of $a^n+b^n$ i get
$2^{3^{n}}+1=(2+1)(2^{3^n-1}-2^{3^n-2}+....+1)$
but how we get above ???

HInt:
Use $$(x+1)(x^2-x+1)=x^3+1$$, successively.
