Fibonacci identity: $f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n$ [duplicate]

We have the following (easily proved by induction):

$\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n = \begin{pmatrix} f_{n+1} & f_n \\ f_n & f_{n-1} \\ \end{pmatrix}$

Equating the determinants of the matrices gives us the identity immediately.

Let us try to find gcd of $F_n$ and $F_{n+1}$ using Extended Euclidean algorithm. I will write the steps algorithm in a table; this table method was also explained in some Bill Dubuque's posts.

$\begin{array}{|l||c|c|} \hline F_{n+1} & 1 & 0 \\\hline F_{n} & 0 & 1 \\\hline F_{n-1} & 1 & -1 \\\hline F_{n-2} & -1 & 2 \\\hline F_{n-3} & 2 & -3 \\\hline \vdots & \vdots & \vdots \\\hline F_{n-k} & (-1)^{k+1}F_k & (-1)^kF_{k+1} \\\hline \end{array} $

After a few steps we can guess the $k$-th line, which gives us the following formula:
$F_{n-k}=(-1)^{k+1}F_kF_{n+1}+(-1)^kF_{k+1}F_n=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)$.

For $k=n-1$ we get Cassini's identity $F_1=(-1)^n(F_{n-1}F_{n+1}-F_n^2)$.

So the only thing we have to do is to verify the above formula, which can be done easily by induction on $k$.

Inductive step: We know that:
$F_{n-k}=(-1)^{k+1}(F_kF_{n+1}-F_{k+1}F_n)=-(-1)^{k}(F_kF_{n+1}-F_{k+1}F_n)$
$F_{n-(k-1)}=(-1)^{k}(F_{k-1}F_{n+1}-F_{k}F_n)$

Since $F_{n-(k+1)}=F_{n-(k-1)}-F_{n-k}$, we get
$F_{n-(k+1)}=(-1)^{k}[(F_{k-1}+F_k)F_{n+1}-(F_k+F_{k+1})F_n]=(-1)^{k+2}(F_{k+1}F_{n+1}-F_{k+2}F_n)$
which completes the inductive step.