factorial - Limit of a function not using Stirling's Approximation

I want to compute the following limit:
$$\lim_{n\to\infty} \frac{\left(\frac{e}{F_{n+1}}\right)^{F_{n+1}} F_{n+1}!}{\left(\frac{e}{F_n}\right)^{F_n} F_n!},$$

where $F_n$ is the $n$th Fibonacci number. The limit is easily computed by using Stirling's approximation $n! \simeq \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$:

$$\lim_{n\to\infty} \frac{\left(\frac{e}{F_{n+1}}\right)^{F_{n+1}} F_{n+1}!}{\left(\frac{e}{F_n}\right)^{F_n} F_n!} = \lim_{n\to\infty} \frac{\left(\frac{e}{F_{n+1}}\right)^{F_{n+1}} \sqrt{2\pi F_{n+1}} \left(\frac{F_{n+1}}{e}\right)^{F_{n+1}}}{\left(\frac{e}{F_n}\right)^{F_n} \sqrt{2\pi F_n} \left(\frac{F_n}{e}\right)^{F_n}}\\=\lim_{n\to\infty}\sqrt{\frac{F_{n+1}}{F_n}}\\=\sqrt{\frac{1+\sqrt{5}}{2}}.$$

Is it possible to show this without using Stirling's approximation?