Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$

Question

Let $g_n= {2^2}^n +1 $. Prove $g_0 · g_1 · · · g_{n−1} = g_{n} − 2$.

I'm not sure how to start this proof, if it should be done algebraically or if I should try to use a proof by induction.

Answer 1

Induction is the way to go. Having established the base case, assume $g_0\cdots g_{n-1}=g_n-2$. Then, $$g_0\cdots g_{n-1}g_n=(g_n-2)g_n=g_n^2-2g_n$$ Now, expand the righthand side and see what you get.