A continued fraction for $\sqrt{2}\Bigg({e^\sqrt{2}-1 \above 1.5pt e^\sqrt{2}+1 }\Bigg)$ This question is about the integer sequence A123168. The title of the sequence reads: Continued fraction for - $$\sqrt{2}\Bigg({e^\sqrt{2}-1 \above 1.5pt e^\sqrt{2}+1 }\Bigg)$$ A comment claims that 'this continued fraction shows that $e^\sqrt{2}$ is irrational". I thought Lindemann & Weierstrass theorem established the irrationality of $e^\sqrt{2}$ so what is the motivation behind this particular continued fraction and the irrationality of $e^\sqrt{2}$? How do you write that continued fraction out?
note: I am not asking for a proof of the irrationality of $e^\sqrt{2}$.
 A: By Gauss' continued fraction,
$$ \tanh(x)=\cfrac{x}{1+\cfrac{x^2}{3+\cfrac{x^2}{5+\cfrac{x^2}{7+\ldots}}}}\tag{1}$$
hence by evaluating at $x=\frac{1}{\sqrt{2}}$,
$$ \sqrt{2}\tanh\left(\frac{1}{\sqrt{2}}\right)=\color{green}{\sqrt{2}\,\frac{e^{\sqrt{2}}-1}{e^{\sqrt{2}}+1}} = \cfrac{1}{1+\cfrac{1/2}{3+\cfrac{1/2}{5+\cfrac{1/2}{7+\ldots}}}}\tag{2}$$
and by converting the RHS of $(2)$ into an ordinary continued fraction,
$$ \color{green}{\sqrt{2}\,\frac{e^{\sqrt{2}}-1}{e^{\sqrt{2}}+1}} = \cfrac{1}{1+\cfrac{1}{6+\cfrac{1}{5+\cfrac{1}{14+\ldots}}}}=\color{green}{[0;1,6,5,14,9,22,13,30,\ldots]}\tag{3}$$
proving that the LHS of $(3)$ is an irrational number (since the continued fraction has an infinite number of terms) and not an algebraic number over $\mathbb{Q}$ with degree $2$ (otherwise, by Lagrange's theorem, the continued fraction would be periodic from some point on). Assuming that $e^{\sqrt{2}}=\frac{p}{q}\in\mathbb{Q}$, we have that the LHS of $(3)$ is a rational multiple of $\sqrt{2}$, hence an algebraic number over $\mathbb{Q}$ with degre $2$. We have already proved that is not our case, hence
$$ (3)\implies \color{red}{e^{\sqrt{2}}\not\in\mathbb{Q}}.\tag{4} $$
