Induction on Fermat Numbers: $F_n = \prod_{j=0}^{n-1}F_j+2$

Question

Is the Following Proof Correct?

Theorem. Given that $\forall n\in\mathbf{N}(F_n = 2^{2^n}+1)$ show that the following is true $$\forall n\in{1,2,3...}\left(F_n = \prod_{j=0}^{n-1}F_j+2\right)$$ Proof. We construct the proof by recourse to Mathematical-Induction.

Basis-Step: For $n=1$ we have $F_1=F_0+2=(2^{2^0}+1)+2 = (2+1)+2=5$.

Inductive-Step: Assume now for an arbitrary $k\in\{1,2,3,...\}$ the following $$F_k = \left(\prod_{j=0}^{k-1}F_j+2\right)\tag{1}$$ consequently using $(1)$ we have $$\left(\prod_{j=0}^{k}F_j+2\right) = \left(\prod_{j=0}^{k-1}F_j\cdot F_k+2\right)=(F_k-2)\cdot F_k=(F_k)^{2}-2F_k+2\tag{2}$$ thus $(F_k)^{2}-2F_k+2 = (2^{2^k}+1)^2-2(2^k+1)+2$ expanding all terms we have $(2^{2^k}+1)^2-2(2^k+1)+2 = 2^{2^{k+1}}+2\cdot 2^{2^k}+1-2\cdot 2^{2^k}-2+2 = 2^{2^{k+1}}+1 = F_{k+1}$ thus $$F_{k+1}=\left(\prod_{j=0}^{k}F_j+2\right)\tag{3}$$

$\blacksquare$

Answer 1

Your proof is fine, but I think it's a little easier if you define the numbers $F_n$ by the recurrence $$F_n=F_0F_1\cdots F_{n-1}+2$$ and then prove the identity $F_n=2^{2^n}+1$ by induction.

Basis step: $$F_0=3=2^{2^0}+1$$ Inductive step: $$F_{n+1}=F_0\cdots F_{n-1}F_n+2$$ $$F_{n+1}-2=(F_0\cdots F_{n-1})F_n=(F_n-2)F_n=\left(2^{2^n}-1\right)\left(2^{2^n}+1\right)=\left(2^{2^n}\right)^2-1=2^{2^{n+1}}-1$$ $$F_{n+1}=2^{2^{n+1}}+1$$

Answer 2

Looks good to me :)

NB - those are Fermat numbers, not Fermat primes.