Fibonacci Numbers proof, circular reasoning? I got this from "Number Theory" by George E. Andrews (1971), at the end of the first chapter, he asks for proofs by mathematical induction about Fibonacci Numbers as exercices. In one of them I am asked to show that 
$$(\forall \, n \in \Bbb Z^+)((F_{n+1})^2-F_nF_{n+2}=(-1)^n)$$
which has already has been asked on Math S.E.: "Fibonacci numbers and proof by induction" but I am not looking for a complete solution here; rather, since I am learning the subject on my own I would appreciate it if someone could simply tell me if my reasoning is correct, it does not "feel" concrete to me. 
Going through the usual steps I first check  that the base case $F_2^2-F_1F_3=1-2=(-1)^1$ is true such that the inductive step can be taken. Then, assuming $(F_{k+1})^2-F_kF_{k+2}=(-1)^k$ is true, I try to show that it implies that $(F_{k+2})^2-F_{k+1}F_{k+3}=(-1)^{k+1}$ is also true. 
Then, doing a bit of algebra I get,
$$\begin{align}
(F_{k+2})^2-F_{k+1}F_{k+3} & = (-1)^{k+1}=(-1)^k(-1)\\
& = ((F_{k+1})^2-F_kF_{k+2})(-1)\\
& = F_kF_{k+2}-(F_{k+1})^2\\
(F_{k+2})^2-F_{k+1}F_{k+3} + (F_{k+1})^2-F_kF_{k+2} & = 0\\
(F_{k+2})^2-F_{k+1}F_{k+3} + (-1)^k & = 0\\
(F_{k+2})^2-F_{k+1}F_{k+3} & = -(-1)^k = (-1)(-1)^k\\
& = (-1)^{k+1}\\
\end{align}$$
$$\tag*{$\blacksquare$}$$
I am not sure about this, it feels like circular reasoning, is it?
Thanks a lot for the valuable input!

Yes, your answer is wrong, because you assume that $(F_{k+2})^2 - F_{k+1}F_{k+3} = (-1)^{k+1}$ but it's what you want to prove. By the way, here is the algebra:
$(F_{k+2})^2 - F_{k+1}F_{k+3} = (F_{k+1}+F_k)^2 - F_{k+1}(F_{k+2} + F_{k+1})= (F_{k+1}^2 +2F_{k+1}F_k +F_k^2) - F_{k+1}(F_{k+1} + F_k + F_{k+1}) = -F_{k+1}^2+F_{k+1}F_k +F_k^2 = -(F_{k+1}^2-F_k(F_{k+1} +F_k))= -(F_{k+1}^2-F_kF_{k+2})= -(-1)^k=(-1)^{k+1} $
