Fibonacci numbers and golden ratio: $\Phi = \lim \sqrt[n]{F_n}$

Let $\Phi$ be the golden ratio and $F_n$ be the usual Fibonacci numbers. How can I derive the following formula?

$$ \Phi = \lim_{n\rightarrow \infty} \sqrt[n]{F_n} $$

I know the usual relation $$ \Phi = \lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n} \quad , $$ and Wikipedia tells me that $$ \Phi^a = \lim_{n\rightarrow \infty} \frac{F_{n+a}}{F_n} \quad . $$

My first idea was to set $a = n$, which gives $$ \Phi = \lim_{n\rightarrow \infty} \sqrt[n]\frac{F_{n+n}}{F_n} \quad , $$

EDIT: We can also do $$ \Phi = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{F_{n+n}}{F_n}\frac{F_n}{F_n}} \quad , $$ but I am totally stuck here...

It is a standard result that $F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}$.

Then $\lim_{n\rightarrow \infty} \sqrt[n]{F_n} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \sqrt[n]{\frac{\phi^n}{\sqrt{5}}} = \lim_{n\rightarrow \infty} \frac{\phi}{\sqrt[n]{\sqrt{5}}} = \phi$ .

We have

$$F_n=\frac1{\sqrt5}\left(\underbrace{\frac{1+\sqrt5}{2}}_{=:\alpha}\right)^n-\frac1{\sqrt5}\left(\underbrace{\frac{1-\sqrt5}{2}}_{=:\beta}\right)^n$$ and since $|\beta|<|\alpha|$ then $$|\beta|^n=_\infty o(|\alpha|^n)$$ hence $$\sqrt[n]{F_n}\sim_\infty \alpha=:\Phi$$