How to factorize $2^{3^{k}}+1$

Question

My question: How do we get $2^{3^{k}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \ldots\left(2^{2 \cdot 3^{k-1}}-2^{3^{k-1}}+1\right)$$(k\ge 0)$

I know $2^{3^{k}}+1=(2+1)(2^{3^{k}-1}-2^{3^{k}-2}+...+1)$

Any help will be appreciated. Thanks:D

Answer 1

$$ 2^{3^k}+1=\left(2^{3^{k-1}}\right)^3+1 =\left(2^{3^{k-1}}+1\right)\left(2^{2 \cdot 3^{k-1}}-2^{3^{k-1}}+1\right) \tag*{(*)} $$ Replacing $k$ by $k-1$ in $(*)$ we have $$2^{3^{k-1}}+1=\left(2^{3^{k-2}}\right)^3+1 =\left(2^{3^{k-2}}+1\right)\left(2^{2 \cdot 3^{k-2}}-2^{3^{k-2}}+1\right) \tag*{(**)}$$ and $$2^{3^k}+1= \left(2^{3^{k-2}}+1\right)\left(2^{2 \cdot 3^{k-2}}-2^{3^{k-2}}+1\right) \left(2^{2 \cdot 3^{k-1}}-2^{3^{k-1}}+1\right) \tag*{(***)} $$ Repeatedly reducing $k$ by $1$, we get $$\boxed{2^{3^{k}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \ldots\left(2^{2 \cdot 3^{k-1}}-2^{3^{k-1}}+1\right)}$$