$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to

$\sqrt{2+\sqrt{3}} \times \sqrt{2+\sqrt{2+\sqrt{3}}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$

Let's define:

$a = \sqrt{3}$

$b = \sqrt{2+a} = \sqrt{2+\sqrt{3}}$

$c = \sqrt{2+b} = \sqrt{2+\sqrt{2+\sqrt{3}}}$

$d = \sqrt{2+c} = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}$

Our expression is then: $b \times c \times d \times \sqrt{2-c}$

The key insight is that these nested square roots form a pattern where:

$b^2 = 2+a$

$c^2 = 2+b$

$d^2 = 2+c$

From this, we get:

$c^2-2 = b$

$d^2-2 = c$

For the term $\sqrt{2-c}$, note that:

$\sqrt{2-c} \times d = \sqrt{2-c} \times \sqrt{2+c} = \sqrt{(2-c)(2+c)} = \sqrt{4-c^2} = \sqrt{4-(2+b)} = \sqrt{2-b}$

Similarly:

$\sqrt{2-b} \times c = \sqrt{2-b} \times \sqrt{2+b} = \sqrt{4-b^2} = \sqrt{4-(2+a)} = \sqrt{2-a}$

And:

$\sqrt{2-a} \times b = \sqrt{2-a} \times \sqrt{2+a} = \sqrt{4-a^2} = \sqrt{4-3} = \sqrt{1} = 1$

Therefore:

$b \times c \times d \times \sqrt{2-c}$

$b \times c \times \sqrt{2-b} = b \times 1 = 1$

$b \times \sqrt{2-a}=1$

So the value of the given expression is $1$.