Cassini's Identity/Proof 2

Proof
First we use this lemma:

Then the determinant of both sides is taken.

The follows directly from the order 2 determinant:
 * $\begin{bmatrix}

F_{n + 1} & F_n     \\ F_n       & F_{n - 1} \end{bmatrix} = F_{n + 1} F_{n - 1} - F_n^2$

Now for the :

Basis for the Induction

 * $\begin{vmatrix}

1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \paren {-1}^1$

Induction Hypothesis
For $k \in \Z_{>0}$, it is assumed that:


 * $\begin{vmatrix}

1 & 1 \\ 1 & 0 \end{vmatrix}^k = \paren {-1}^k$

It remains to be shown that:


 * $\begin{vmatrix}

1 & 1 \\ 1 & 0 \end{vmatrix}^{k + 1} = \paren {-1}^{k + 1}$

Induction Step
The induction step follows from Determinant of Matrix Product:


 * $\begin{vmatrix}

1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = \paren {-1}^k \paren {-1} = \paren {-1}^{k + 1}$

Hence by induction:


 * $\forall n \in \Z_{>0}: \begin{vmatrix}

1 & 1 \\ 1 & 0 \end{vmatrix}^n = \paren {-1}^n$