Take $n = 0$ and see that
$$\begin{bmatrix} f(n) \\ f(n+1) \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$
is clearly true: We just see the first two terms.
Suppose for a particular $n \in \mathbb{N}$,
$$\begin{bmatrix} f(n) \\ f(n+1) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^n\begin{bmatrix} 0 \\ 1 \end{bmatrix}$$
then, $$\begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} f(n) \\ f(n+1) \end{bmatrix} = \begin{bmatrix} f(n+1) \\ f(n) + f(n+1) \end{bmatrix} = \begin{bmatrix} f(n+1) \\ f(n+2) \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix}^{n+1}\begin{bmatrix} 0 \\ 1 \end{bmatrix}$$