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 
$$F_{k+1}=\left(\prod_{j=0}^{k}F_j+2\right)\tag{3}$$
$\blacksquare$

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