It is well known that $$\operatorname{R}(-e^{-\pi})=-\cfrac{e^{-\frac{\pi}{5}}}{1-\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1-\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}-\sqrt{\frac{5-\sqrt{5}}{2}}$$ where $\operatorname{R}$ is the Rogers-Ramanujan continued fraction: $$\operatorname{R}(q)=\cfrac{q^{\frac{1}{5}}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}},\, q=e^{\pi i\tau}.$$

But I'm interested in $\operatorname{R}(e^{-\pi})$. Numerically, I checked that it agrees with the root of the following octic equation near $x=\frac{1}{2}$ to at least $16$ decimal places: $$x^8+14x^7+22x^6+22x^5+30x^4-22 x^3+22 x^2-14x+1=0;$$ the root turns out to be equal to $$\frac{\sqrt{5}-1}{2}\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}.$$

So is it true that $$\frac{e^{-\frac{\pi}{5}}}{1+\cfrac{e^{-\pi}}{1+\cfrac{e^{-2\pi}}{1+\cfrac{e^{-3\pi}}{1+\ddots}}}}=\frac{\sqrt{5}-1}{2}\frac{\sqrt[4]{5}+\sqrt{2+\sqrt{5}}}{\sqrt{5}+\sqrt{2+\sqrt{5}}}?$$ The $2$'s and $5$'s under the square roots seem very suggestive of the nature of $\operatorname{R}$.

Also, how could it be proved in that case?

Using $$\frac{1}{\operatorname{R}(q)}-\operatorname{R}(q)=\frac{\left(q^{\frac{1}{5}};q^{\frac{1}{5}}\right)_{\infty}}{q^{\frac{1}{5}}(q^5;q^5)_{\infty}}+1$$ and $$\frac{\eta (e^{-\pi\sqrt{n}})}{\eta \left(e^{-\frac{\pi}{\sqrt{n}}}\right)}=n^{-\frac{1}{4}},\, n\gt 0$$ (where $\eta (q)=q^{\frac{1}{12}}\prod_{n\ge 1}(1-q^{2n})$ is the Dedekind eta function), I've been able to evaluate $\operatorname{R}(-e^{-\pi})$ and $\operatorname{R}(e^{-2\pi})$, but I don't know how could it be used to evaluate $\operatorname{R}(e^{-\pi})$. Perhaps something else is necessary.

I was inspired by Ramanujan's first letter to Hardy, where Ramanujan's $7\text{th}$ theorem states that $$\cfrac{1}{1+\cfrac{e^{-\pi\sqrt{n}}}{1+\cfrac{e^{-2\pi\sqrt{n}}}{1+\cfrac{e^{-3\pi\sqrt{n}}}{1+\ddots}}}}$$ can be exactly found for any $n\in\mathbb{Q}^{+}$.