Prove that $2-\cfrac{\pi^2}{6-\cfrac{\pi^2}{10-\cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0$

Prove that

$$2 - \cfrac{\pi^2}{6-\cfrac{\pi^2}{10 - \cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0$$

My thoughts: One common approahch is setting $x = \text{LHS}$ and we express LHS as $x = \frac{\pi^2}{f(x)}$. But it's hard to translate the denominator as a function of $x$

$\pi^2$ also reminds everybody of the basel problem $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ but what can we do about it?

The standard reference for continued fractions (at least as of 100 years ago) is

Perron, Oskar, Die Lehre von den Kettenbrüchen. Leipzig—Berlin: B. G. Teubner. xiii, 520 S. $(8^\circ)$ (1913). ZBL43.0283.04.

Section 81, Satz 3 states (in modern language): Consider the continued fraction $$ b_0 + \frac{a_1}{b_1 + \displaystyle\frac{a_2}{b_2+\ddots}} $$ with $a_n = a$ and $b_n = dn+c$. Provided $a \ne 0, c \ne 0,$ and $d \ne 0$, the value of the continued fraction is $$ V = \frac{c \; {}_0F_1(;c/d;a/d^2)}{\;{}_0F_1(;1+c/d;a/d^2)\;} $$

Take $a=-\pi^2$ so $a_n = -\pi^2$ and
$c=2, d=4$ so $b_n = 4n+2$.
The value of the continued fraction is computed as $$ V = \frac{\;2\;{}_0F_1(;1/2;-\pi^2/16)\;}{{}_0F_1(;3/2;-\pi^2/16)} $$ But the numerator is $2\cos(2\sqrt{\pi^2/16}\;) = 0$ and the denominator is $2/\pi$, so $$ V = 0 . $$

Another answer.
HERE we find the formula

$$ {\displaystyle e^{\frac {x}{y}}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}} }$$

Substitute $x=i\pi, y=1$ $$ {\displaystyle e^{i\pi}=1+{\cfrac {2i\pi}{2 -i\pi+{\cfrac {-\pi^{2}}{6 +{\cfrac {-\pi^{2}}{10 +{\cfrac {-\pi^{2}}{14 +{\cfrac {-\pi^{2}}{18 +\ddots }}}}}}}}}}} $$ We recognize most of the required terms inside this. Rearrange this (using $e^{i\pi}=-1$) to get $$ 2 - \cfrac{\pi^2}{6-\cfrac{\pi^2}{10 - \cfrac{\pi^2}{14-\cfrac{\pi^2}{...}}}} = 0 $$