<s>[INST] Induction on Fermat Numbers: $F_n = \prod_{j=0}^{n-1}F_j+2$ 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 
$$F_{k+1}=\left(\prod_{j=0}^{k}F_j+2\right)\tag{3}$$
$\blacksquare$
 [/INST] Looks good to me :)
NB - those are Fermat numbers, not Fermat primes.
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