There are different ways of generating $\phi=\frac{1+\sqrt{5}}{2}= 1.618$, and one of them is presented in Professor Strang's book Introduction to Linear Algebra. I'm modifying a bit the details, but the idea is what counts:
In the Fibonacci series, the recursive formula is such that it can be expressed as:
$\begin{bmatrix}F_{n+1}\\F_{n}\end{bmatrix}=\color{orange}{\begin{bmatrix}1&1\\1&0\end{bmatrix}}\begin{bmatrix}F_{n}\\F_{n-1}\end{bmatrix}$
Starting at $u_0=\begin{bmatrix}F_1\\F_0\end{bmatrix}=\color{brown}{\begin{bmatrix}1\\0\end{bmatrix}}$, we can proceed as
$\begin{bmatrix}F_2\\F_1\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}^1\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}$
and matrix multiply $100$ times to get $F_{100}$:
$\begin{bmatrix}F_{101}\\\color{blue}{F_{100}}\end{bmatrix}=\color{orange}{\begin{bmatrix}1&1\\1&0\end{bmatrix}}^{\color{blue}{100}}\color{brown}{\begin{bmatrix}1\\0\end{bmatrix}}$
which is straightforward if we can eigendecompose the matrix $\color{orange}{\begin{bmatrix}1&1\\1&0\end{bmatrix}}$:
$\color{orange}{\begin{bmatrix}1&1\\1&0\end{bmatrix}}=S\Lambda S^{-1}=\begin{bmatrix}\vert&\vert\\\text{ev}_1&\text{ev}_2\\\vert&\vert\end{bmatrix}\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\begin{bmatrix}\vert&\vert\\\text{ev}_1&\text{ev}_2\\\vert&\vert\end{bmatrix}^{-1}$
In which case:
$\begin{bmatrix}F_{101}\\\color{blue}{F_{100}}\end{bmatrix}=\begin{bmatrix}\vert&\vert\\\text{ev}_1&\text{ev}_2\\\vert&\vert\end{bmatrix}\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}^{\color{blue}{100}}\begin{bmatrix}\vert&\vert\\\text{ev}_1&\text{ev}_2\\\vert&\vert\end{bmatrix}^{-1}\color{brown}{\begin{bmatrix}1\\0\end{bmatrix}}\tag 1$
So we have to find the eigenvalues of $\begin{bmatrix}1&1\\1&0\end{bmatrix}$:
$0=\begin{bmatrix}1&1\\1&0\end{bmatrix}-\lambda\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}1-\lambda&1\\1&-\lambda\end{bmatrix}$
Solving for the eigenvalues results in $\large\color{red}{\lambda_1=\frac{1+\sqrt{5}}{2}}$ and $\large\lambda_2=\frac{1-\sqrt{5}}{2}.$
So we have "generated" the golden ratio as the first eigenvalue of the matrix expressing the transformation from each column vector containing two adjacent Fibonacci numbers to to the next overlapping ordered pair.
Here is the plot in relation to the basis vectors in the transformation $\small\begin{bmatrix}\color{orange}{1}&\color{red}{1}\\\color{orange}{1}&\color{red}{0}\end{bmatrix}:$

Just to wrap up the story, the eigenvectors are:
$\text{ev}_1=\begin{bmatrix}\lambda_1\\1\end{bmatrix}$ and $\text{ev}_2=\begin{bmatrix}\lambda_2\\1\end{bmatrix}$
and going back to Eq.1:
$\begin{bmatrix}F_{101}\\\color{blue}{F_{100}}\end{bmatrix}=\frac{1}{\lambda_1-\lambda_2}\begin{bmatrix}\lambda_1&\lambda_2\\1&1\end{bmatrix}\begin{bmatrix}\lambda_1^{100}&0\\0&\lambda_2^{100}\end{bmatrix}\begin{bmatrix}1&-1\\-\lambda_2&\lambda_1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}$
and since $\lambda_2^{100}\simeq 0,$
$\begin{bmatrix}F_{101}\\\color{blue}{F_{100}}\end{bmatrix}\simeq\frac{1}{\lambda_1-\lambda_2}\begin{bmatrix}\lambda_1&\lambda_2\\1&1\end{bmatrix}\begin{bmatrix}\lambda_1^{100}\\0\end{bmatrix}=\frac{1}{\lambda_1-\lambda_2}\begin{bmatrix}\lambda_1^{101}\\\lambda_1^{100}\end{bmatrix}$
and
$F_{100}\simeq\frac{\lambda_1^{100}}{\lambda_1-\lambda_2}=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{100}$
which is the approximation in the OP:
$\color{blue}{F_n\simeq \frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n}}$