Nested Radicals from Brazil Show that
$$
\frac{ \sqrt{2} }{ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} - \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1}} = \sqrt[8]{ 1 + 2 \sqrt{ \sqrt{5} -2 } }.
$$
What I've tried so far.
$$
\begin{align*}
E & = \frac{ \sqrt{2} }{ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} - \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1}} \\
& = \frac{1}{\sqrt{2}} \cdot \left[ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} + \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1} \right] \\
& = \frac{1}{\sqrt{2}} \cdot \left[ \sqrt{ a + 1 } + \sqrt{a -1} \right]
\end{align*}
$$
Squaring $E$,
$$
\begin{align*}
E^2 & = \frac{1}{2} \left[ a + 1 + a - 1 + 2\sqrt{a+1} \sqrt{a-1} \right] \\
& = \frac{1}{2} \left[ 2a + 2 \sqrt{a^2-1} \right] \\
& = a + \sqrt{a^2-1}
\end{align*}
$$
Then, isolating the radical
$$
\begin{align*}
E^2 - a & = \sqrt{a^2-1} \\
E^4 + a^2 - 2aE^2 & = a^2 - 1
\end{align*}
$$
We obtain $a$,
$$
a = \frac{1}{2} \left( E^2 + \frac{1}{E^2} \right).
$$
And then, nothing useful comes up.

With your notations:
$E^2 = a +\sqrt{a^2-1}\\
E^8 =a^4+4a^3 \sqrt{a^2-1} +6a^2(a^2-1)+4a(a^2-1)\sqrt{a^2-1}+(a^2-1)^2\\
E^8 = 8a^4-8a^2+1+(8a^3-4a)\sqrt{a^2-1}$

We can write that:
$2\sqrt{\sqrt{5}-2} = \dfrac{2  \sqrt{\sqrt{5}-2} \sqrt{\dfrac{\sqrt{5}+2}{4}}} {\sqrt{\dfrac{\sqrt{5}+2}{4}}} =\dfrac{1}{a^2}  $

So, we want to show that    $\ E^8 = 1+\dfrac{1}{a^2}$
We want to show that   $\ 8a^4-8a^2 + (8a^3-4a)\sqrt{a^2-1}=\dfrac{1}{a^2}$
We want to show that  $\ (8a^5-4a^3)\sqrt{a^2-1} = 1-8a^6+8a^4$

Let $\ x=(8a^5-4a^3)\sqrt{a^2-1}\ $ and $\ y=1-8a^6+8a^4\ $. It's easy to check that $x$ and $y$ are positive.
$x^2-y^2 = (8a^5-4a^3)^2 (a^2-1) - (1-8a^6+8a^4)^2\\
x^2-y^2 = (64a^{10}-64a^8+16a^6)(a^2-1)-(64a^{12}-128a^{10}+64a^8-16a^6+16a^4+1)\\
x^2-y^2 = 16a^8-16a^4-1$
And, now, it's easy to check that $x=y$.
