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I know that this should be a proof by induction, and my hunch is strong induction, but I am unsure about how to go about this.

In my base case, I made sure that the proposition held for n = 0, n = 1, and n = 2 before moving onto my inductive step.

Am I trying to show that $2^{2^{(n+1)}} + 1$ = ($2^{2^0}+ 1 * ... *2^{2^{n-1}}+1*2^{2^n}+1)+2 $?

Tom Brady
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  • I think this has been asked here before. What should help if we denote $F(n):=2^{2^n}+1$ : $$F_{n+1}=(F_n-1)^2+1$$ – Peter Oct 30 '22 at 20:04
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    Going forward, for generic problems of this type, the first step is $~\color{red}{\text{Elbow Grease}},~$ using scratch paper. For example, with this problem, you (for example) let $n$ be each element in ${1,2,3,4,5}$. For each such element, you compute $F_n$ both by the formula and by the assertion. If they match in each case, expand your intuition by examining the data to determine why they match. Then, you are ready to try a proof, perhaps by some form of induction. – user2661923 Oct 30 '22 at 20:32

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