For $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$. Show that $g_0\cdot g_1\cdots g_{n-1} = g_n -2$.

Question

For $n \in Z^{\geq 0}$, define $g_n = 2^{2^n} + 1$. Show that $g_0\cdot g_1\cdots g_{n-1} = g_n -2$ for all $n \in Z^+$.

I thought that this could be proved using induction, but then the base case wouldn't work since for n=0, there would be nothing on the left hand side of the equation. Is there a way to prove this algebraically or otherwise?

Answer 1

The base case, for $n=1$, is $g_0 = g_1-2$ or $2^1+1 = 2^2+1-2$ or $3 = 5-2$.

From this, you can use induction.