Rationalize the denominator of the following fraction:

$$
\frac{1}{\sqrt{(2+\sqrt{3})(2+\sqrt{2+\sqrt{3}})(2+\sqrt{2+\sqrt{2+\sqrt{3}}})}}
$$
Let's denote the three factors under the square root as $a, b, c$ respectively. By taking the second term in these binomial sums with a negative sign, we obtain the differences $a^{\prime}, b^{\prime}, c^{\prime}$. Accordingly,

$$
\begin{aligned}
& c \cdot c^{\prime}=2^{2}-(2+\sqrt{2+\sqrt{3}})=2-\sqrt{2+\sqrt{3}}=b^{\prime} \\
& b \cdot b^{\prime}=2^{2}-(2+\sqrt{3})=2-\sqrt{3}=a^{\prime} \\
& a \cdot a^{\prime}=2^{2}-3=1
\end{aligned}
$$

Extend the given expression by $\sqrt{c^{\prime}}$, and pay attention to the identities labeled (1), (2), (3):

$$
\frac{1}{\sqrt{a b c}}=\frac{\sqrt{c^{\prime}}}{\sqrt{a b c c^{\prime}}}=\frac{\sqrt{c^{\prime}}}{\sqrt{a b b^{\prime}}}=\frac{\sqrt{c^{\prime}}}{\sqrt{a a^{\prime}}}=\sqrt{c^{\prime}}
$$

Thus,

$$
\frac{1}{\sqrt{(2+\sqrt{3})(2+\sqrt{2+\sqrt{3}})(2+\sqrt{(2+\sqrt{2+\sqrt{3}})}}}=\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}
$$