I was working with continued fractions and the Fibonacci sequence. And concluded this.
Firstly, we know $$\lim_{n \to \infty}\frac{F_{n+1}}{F_{n}}= \phi$$ and easily you could prove it is using its recursive formula by induction and some manipulation to the equation. For that, $$\lim_{n\to \infty}\frac{F_{n+1}^{k}}{F_{n}^{k}}= \phi^{k}$$ And we could express the odd powers of the golden ratio with a continued fraction that has Lucas numbers as $$\phi^{n}=[ L_{n};\overline{L_{n}}]$$ and of course if n is odd. Finally, there is a unique identity that connects Fibonacci and Lucas numbers, which is $$F_{n+k}=L_{k}F_{n}+(-1)^{k-1}F_{n-k}$$ Let us divide with $F_{n}$ in each hand $$\frac{F_{n+k}}{F_n}=L_{k}+\frac{(-1)^{k-1}F_{n-k}}{F_{n}}$$ Assuming $k$ is odd, then $$\frac{F_{n+k}}{F_n}=L_{k}+\frac{F_{n-k}}{F_{n}}$$ that we could just write like $$\frac{F_{n+k}}{F_n}=L_{k}+\frac{1}{\dfrac{F_{n}}{F_{n-k}}}$$ and you could set $n$ to be anything, like let $n=r+k$. Which makes it $$\frac{F_{n+k}}{F_n}=L_{k}+\frac{1}{\dfrac{F_{r+k}}{F_{r}}}$$ and making the same step in $F_{r+k}$ we get $$\frac{F_{n+k}}{F_n}=L_{k}+\frac{1}{L_{k}+\dfrac{1}{\dfrac{F_{r}}{F_{r-k}}}}$$ and doing over and over for such $\infty$, we concluded $$\frac{F_{n+k}}{F_n}=L_{k}+\cfrac{1}{L_{k}+\cfrac{1}{L_{k}+\cfrac{1}{L_{k}+\ddots}}}=[ L_{k};\overline{L_{k}}]$$ then it should be for all odd $k$'s.$$\lim_{n\to \infty}\frac{F_{n+1}^{k}}{F_{n}^{k}}=\lim_{n\to \infty}\frac{F_{n+k}}{F_n}=\phi^{k},$$ is true.
However, I think it is hard to be true. It looks fabulous, but I feel the over-'n-over step is not a rigorous thing to do to prove.
Is it true? Is there another precise way to prove it or disprove it?
Let $$g(n,k) = \frac{F_{n+k}}{F_n}. \tag{1}$$ Then if we have established that $$\lim_{n \to \infty} g(n,1) = \phi, \tag{2}$$ then it immediately follows that for any fixed positive integer $k$, $$\lim_{n \to \infty} g(n,1)^k = \left(\lim_{n \to \infty} g(n,1)\right)^k = \phi^k. \tag{3}$$ That is to say, the interchange of the limit and raising to the $k^{\rm th}$ power is justified because the limit $(2)$ exists and is finite.
It is also not difficult to see that $$\frac{F_{n+k}}{F_n} = \frac{F_{n+1}}{F_n} \cdot \frac{F_{n+2}}{F_{n+1}} \cdot \cdots \cdot \frac{F_{n+k}}{F_{n+k-1}}; \tag{4}$$ i.e. the product telescopes: $$g(n,k) = \prod_{j=0}^{k-1} g(n+j,1). \tag{5}$$ And since each factor in the finite product is also finite, $$\lim_{n \to \infty} g(n,k) = \lim_{n \to \infty} \prod_{j=0}^{k-1} g(n+j,1) = \prod_{j=0}^{k-1} \lim_{n \to \infty} g(n+j,1) = \prod_{j=0}^{k-1} \phi = \phi^k. \tag{6}$$