As I previously explained, I am a "hobbyist" mathematician (see here or here); I enjoy discovering continued fractions by using various algorithms I have been creating for several years. This morning, I got some special cases for a rather heavy expression I was working on, finding the beautiful identities: $$ \phi^\phi \;=\; 2 + \displaystyle\frac{2\left(1-1/\phi\right)/\phi} {2\left(2-1/\phi\right) + \displaystyle\frac{3\left(1-2/\phi\right)/\phi} {3\left(2-1/\phi\right) + \displaystyle\frac{4\left(1-3/\phi\right)/\phi} {4\left(2-1/\phi\right) + \displaystyle\frac{5\left(1-4/\phi\right)/\phi} {5\left(2-1/\phi\right) + \;\ddots }}}} $$ Equivalently $$ \phi^\phi \;=\; 2 + \displaystyle\frac{2/\phi\left(\phi-1\right)} {2\sqrt5 + \displaystyle\frac{3\left(\phi-2\right)} {3\sqrt5 + \displaystyle\frac{4\left(\phi-3\right)} {4\sqrt5 + \displaystyle\frac{5\left(\phi-4\right)} {5\sqrt5 + \;\ddots }}}} $$ where $\phi$ is the golden ratio.

I was very excited to have found this $\phi^\phi$ formula and **my question is about how original it is**: did someone already see it or could I be confident in assuming it is new?
Similarly
$$
\phi^{2/\phi}
\;=\;
2 +
\displaystyle\frac{2\left(1-2/\phi\right)}
{2\phi +
\displaystyle\frac{3\left(1-3/\phi\right)}
{3\phi +
\displaystyle\frac{4\left(1-4/\phi\right)}
{4\phi +
\displaystyle\frac{5\left(1-5/\phi\right)}
{5\phi + \;\ddots }}}}
$$
By using a more compact notation (see at the end of this message for an explanation about how to read it), other identities for $\sqrt{3}$, $\sqrt[3]{4}$, $\sqrt[4]{5}$, etc. would be:

$$ \sqrt[a]{a+1} \;=\;2+ \displaystyle\operatorname*{K}_{n=1}^{\infty} \frac {\left(n+1\right)\left(1-an\right)/\left(a+1\right)} {\left(n+1\right)\left(1+a/\left(a+1\right)\right)} $$

These identities are special case of the following conjectured identity (of course, when asking whether the previous formulae are known or not, it also means the formula below should be new also): $$ \left(\displaystyle\frac{x}{x-1}\right)^{x-1} \;=\;2+ \displaystyle\operatorname*{K}_{n=1}^{\infty} \frac {\left(n+1\right)\left(x-n-1\right)/x} {\left(n+1\right)\left(x+1\right)/x} \tag{1}\label{1} $$ that I found from a more general conjecture being related to the following continued fraction: $$ g\left(k,x\right) = \displaystyle\operatorname*{K}_{n=1}^{\infty} \frac {\left(n+1\right)\left(k-n-k/x\right)/x} {\left(n+1\right)\left(1+1/x\right)} $$ for which I empirically noticed the relation (with $\textrm{B}$ the beta function): $$ \begin{array}{l} 2\;+\; g\left(\alpha,\xi\right) \;+\; g\left(\alpha,\displaystyle\frac{\xi}{\xi-1}\right) \\[8pt] \qquad\qquad =\; \alpha\left(\xi-1\right)^{\alpha/\xi-1}\left(\displaystyle\frac{\xi}{\xi-1}\right)^{\alpha-2} \textrm{B}\left(\alpha/\xi, \alpha-\alpha/\xi\right) \end{array} \tag{2}\label{2} $$ where it is easy to notice that the case $\alpha=\xi/\left(\xi-1\right)$ makes $g\left(\alpha,\xi\right)=0$ leading to an identity containing a single continued fraction (which can later be simplified as above).

The previous notation is the one I use; I find it convenient and it can be found for instance in Continued Fractions with Applications by Lorentzen & Waadeland, but I know that some people don't like it; it has to be read the following way: $$ a_0 + \operatorname*{K}_{n=1}^{\infty} \frac{b_n}{a_n} = a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2}{a_2 + \cfrac{b_3}{a_3 + \ddots}}} $$