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.

Let $x = 4-3\sqrt[4]{5}+2\sqrt[4]{25}-\sqrt[4]{125}$. Then, we have: 
(1) $x = 4 - 3 \cdot 5^{1/4} + 2 \cdot 5^{2/4} - 5^{3/4}$
(2) $5^{1/4}x = 4\cdot 5^{1/4} - 3 \cdot 5^{2/4} + 2 \cdot 5^{3/4} - 5$ (Multiply (1) by $5^{1/4}$)
(3) $(5^{1/4}+1)x = -1 + 5^{1/4} - 5^{2/4} +5^{3/4}$ (Add (1) and (2))
(4) $5^{1/4}(5^{1/4}+1)x = -5^{1/4} + 5^{2/4} - 5^{3/4} +5$ (Multiply (3) by $5^{1/4}$)
(5) $(5^{1/4}+1)^2x = 4$ (Add (3) and (4))
Therefore, $x = \dfrac{4}{(5^{1/4}+1)^2}$. Hence, $\left(\dfrac{2}{\sqrt{x}}-1\right)^4 = \left(\dfrac{2}{\tfrac{2}{5^{1/4}+1}}-1\right)^4 = (5^{1/4}+1-1)^4 = 5$. 
