The proof proceeds by induction. For all $n \in \Z_{\ge 2}$, let $P \left({n}\right)$ be the proposition: :$D_n = F_{n + 1}$ === Basis for the Induction === $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = D_2       | r = \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}       | c =  }} {{eqn | r = \left({1 \times 1}\right) - \left({-1 \times 1}\right)       | c = {{Defof|Determinant of Order 2}} }} {{eqn | r = 2       | c =  }} {{eqn | r = F_3       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} Thus $P \left({2}\right)$ is seen to hold. $P \left({3}\right)$ is the case: {{begin-eqn}} {{eqn | l = D_3       | r = \begin{vmatrix} 1 & -1 & 0 \\ 1 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix}       | c =  }} {{eqn | r = \left({1 \times \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} }\right) - \left({\left({-1}\right) \times \begin{vmatrix} 1 & -1 \\ 0 & 1 \end{vmatrix} }\right) + \left({0 \times \begin{vmatrix} 1 & 1 \\ 0 & 1 \end{vmatrix} }\right)       | c = Expansion Theorem for Determinants: expanding by $1$st row }} {{eqn | r = 1 \times D_2 + 1 \times \left({1 \times 1}\right)       | c = {{Defof|Determinant of Order 2}} }} {{eqn | r = 2 + 1       | c = from above }} {{eqn | r = F_3 + F_2       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_4       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} Thus $P \left({3}\right)$ is seen to hold. $P \left({2}\right)$ and $P \left({3}\right)$ together form the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ and $P \left({k - 1}\right)$ are true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: {{begin-eqn}} {{eqn | l = D_k       | r = F_{k + 1} }} {{eqn | l = D_{k - 1}       | r = F_k }} {{end-eqn}} from which it is to be shown that: :$D_{k + 1} = F_{k + 2}$ === Induction Step === This is the induction step: Recall the definition of $D_n$: :$D_{k + 1} = \underbrace {\begin{vmatrix} 1 & -1 & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \cdots & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix} }_{k + 1 \text { columns} }$ Expanding by $1$st row using the Expansion Theorem for Determinants requires the evaluation of $2$ cofactors: :$1 \times C_{1 1}$ and: :$\left({-1}\right) \times C_{1 2}$ where $C_{r s}$ denotes the determinant of order $k$ obtained from $D_{k + 1}$ by deleting row $r$ and column $s$. By the construction of $D_{k + 1}$, it can be seen that $C_{1 1}$ is $D_k$. The structure of $C_{1 2}$ is seen to be: :$C_{1 2} = \left({-1}\right) \times \underbrace {\begin{vmatrix} 1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix} }_{k \text { columns} }$ By Determinant with Unit Element in Otherwise Zero Column: :$C_{1 2} = \left({-1}\right) \times \underbrace {\begin{vmatrix} 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\ 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\ 0 & 1 & 1 & \cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & -1 & 0 \\ 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\ 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ \end{vmatrix} }_{k - 1 \text { columns} }$ which equals $-D_{k - 1}$. Putting this together, we have: {{begin-eqn}} {{eqn | l = D_{k + 1}       | r = 1 \times C_{1 1} + \left({-1}\right) \times C_{1 2}       | c = Expansion Theorem for Determinants }} {{eqn | r = C_{1 1} - C_{1 2}       | c =  }} {{eqn | r = D_k - \left({-D_{k - 1} }\right)       | c =  }} {{eqn | r = D_k + D_{k - 1}       | c =  }} {{eqn | r = F_{k + 1} + F_k       | c = Induction Hypothesis }} {{eqn | r = F_{k + 2}       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $P \left({k}\right) \land P \left({k - 1}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 2}: D_n = F_{n + 1}$ {{qed}}