How to prove: $\left(\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}}}-1\right)^{4}=5$? Question:

show that: the beautiful ${\tt sqrt}$-identity:
$$
\left({2 \over \sqrt{\vphantom{\Large A}\, 4\ -\ 3\,\sqrt[4]{\,5\,}\
+\ 2\,\sqrt[4]{\,25\,}\ - \,\sqrt[4]{\,125\,}\,}\,}\ -\ 1\right)^{4}
=5
$$

Can you someone have methods to prove this by hand? (Maybe this problem have many methods?because this result is integer. It's a surprise to me.)
Thank you 
Because I found this
$$4\ -\ 3\sqrt[4]{\,5\,}\ +\ 2\sqrt[4]{\,25\,}\ -\ \sqrt[4]{\,125\,}$$ is not square numbers.

[This is a paraphrase on JimmyK4542's elegant answer earlier]
Let $$b=-5^{\frac 14}$$
Then the long expression under the long square root sign becomes an arithmetico-geometric series:
$$k=4+3b+2b^2+b^3$$
Multiplying by $b$:
$$\begin{align}
b\cdot k&=\quad \qquad 4b+3b^2+2b^3+b^4\\
&=\quad \qquad 4b+3b^2+2b^3+5 \end{align}$$
Subtracting:
$$\begin{align}
(b-1)k&=1+b+b^2+b^3\\
&=\dfrac{b^4-1}{b-1}\\
&=\dfrac 4{b-1}\\
k&=\dfrac 4{(b-1)^2}\\
\sqrt{k}&=\dfrac 2{1-b}\\
\left( \dfrac 2{\sqrt{k}} -1\right)^4&={(-b)}^4=5
\end{align}$$
