Once you have the formula, it's routine (if tedious) to check it by expanding out. Finding a nice formula is more challenging. In Maple, you could denest the left side of (1) as follows:
Q:= sqrt((1+4^(1/5))/5^(1/5));
convert(simplify(convert(Q,RootOf)),radical);
$$ \frac{1}{10}\,{5}^{2/5} \left( {4}^{4/5}+{4}^{3/5}+2\cdot {4}^{2/5}-2 \right) $$
EDIT:
Now let's try some reverse engineering. Say you wanted to find a nice formula for a square root where the numerator of the right side involved some linear combination of fifth roots of powers of $2$ with small integer coefficients. We might look at
$$R = a_0 + 2^{1/5} a_1 + 2^{2/5} a_2 + 2^{3/5} a_3 + 2^{4/5} a_4$$
We can square this and extract the coefficients of $2^{i/5}$, $i=0\ldots 4$:
$R^2 = \sum_{i=0}^4 c_i 2^{i/5}$ where
$$ \eqalign{c_0 &= {a_{{0}}}^{2}+4\,a_{{1}}a_{{4}}+4\,a_{{2}}a_{{3}}\cr
c_1 &= 2\,a_{{0}}a_{{1}}+4\,a_{{2}}a_{{4}}+2\,{a_{{3}}}^{2}\cr
c_2 &= 2\,a_{{0}}a_{{2}}+{a_{{1}}}^{2}+4\,a_{{3}}a_{{4}}\cr
c_3 &= 2\,a_{{0}}a_{{3}}+2\,a_{{1}}a_{{2}}+2\,{a_{{4}}}^{2}\cr
c_4 &= 2\,a_{{0}}a_{{4}}+2\,a_{{1}}a_{{3}}+{a_{{2}}}^{2}
}$$
We want most of these (say at least $3$ of the $5$) to be $0$. It's not easy to solve such a system of Diophantine equations, but we can resort to brute force: try all cases where $a_i \in \{-2,-1,0,1,2\}$. In Maple it takes practically no time. Thus one such case is Ramanujan's
$$ (a_0, \ldots, a_4) = (-1,1,0,1,1)$$
which makes
$$ (c_0, \ldots, c_4) = (5,0,5,0,0)$$
Another is
$$ (a_0, \ldots, a_4) = (2, 0, 2, 2, -1)$$
which makes
$$ (c_0, \ldots, c_4) = (20,0,0,10,0) $$
i.e. $$ 20 + 10 \cdot 2^{3/5} = (2 + 2 \cdot 2^{2/5} + 2 \cdot 2^{3/5} - 2^{4/5})^2 $$
Divide by $100$ and take square roots: after checking the sign is right, this says
$$ \sqrt{\dfrac{1 + 2^{-2/5}}{5}} = \dfrac{1 + 2^{2/5} + 2^{3/5} - 2^{-1/5}}{5}$$
I don't know if it's as nice as Ramanujan's formula, but I like it.
Or maybe you'd prefer
$$ \sqrt{\dfrac{4\cdot 2^{3/5} - 3 \cdot 2^{2/5}}{5}} = \dfrac{4 - 2^{1/5} - 2 \cdot 2^{2/5} + 2\cdot 2^{3/5}}{5} $$
or
$$ \sqrt{8+5 \cdot 3^{1/6}+3^{1/2}} = \frac{1+2 \cdot 3^{1/6}+3^{1/3} - 3^{1/2} + 3^{5/6}}{\sqrt{2}}$$
