= Solution
The <diagonalization> gives $A=SDS^{-1}$ and hence $A^{n-1}=SD^{n-1}S^{-1}$: the intervening factors $S^{-1}S$ cancel. The <diagonal matrix> power has entries $\alpha^{n-1},\beta^{n-1}$. Since $1-\beta=\alpha$ and $\alpha-1=-\beta$,
$$
S^{-1}\begin{pmatrix}1\\1\end{pmatrix}=\frac1{\sqrt5}\begin{pmatrix}\alpha\\-\beta\end{pmatrix},\qquad
V_n=\frac1{\sqrt5}\begin{pmatrix}\alpha^{n+1}-\beta^{n+1}\\\alpha^n-\beta^n\end{pmatrix}.
$$
Reading the second component gives the <Binet formula>
$$
\boxed{U_n=\frac{\left((1+\sqrt5)/2\right)^n-\left((1-\sqrt5)/2\right)^n}{\sqrt5}.}
$$
The derivation includes $n=1$, and the expression gives $U_1=U_2=1$, consistent with the specified indexing.
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