Why is $\sqrt{2\sqrt{2\sqrt{2\cdots}}} = 2$? Why is $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\cdots}}}}}}$ equal to 2? Does this work for other numbers?
 A: The number in question is simply
$$2^{1/2+1/4+1/8+\cdots} = 2^1 = 2$$
Yes, this works for other numbers.  More interesting is if the numbers are not equal inside the radicals.  For example, say you have a positive sequence element $a_n$ inside the $n$th radical.  If we assume the expression converges to some value $P$, then
$$\log{P} = \sum_{k=1}^{\infty} \frac{\log{a_n}}{2^n}$$
A: In the case of $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$, this is a limit of the short-chord of a polygon $2^n \cdot x$ as $n$ goes large.  The short-chord is the chord of a regular polygon that spans two edges.  There is a triangle formed by two edges and the short-chord.  As the polygon gets more sides, this triangle approaches a straight line, and the short-chord approaches two edges in length.
A: A general way to ascribe a value to this non rigorously defined formula and to many others similar to it, is to consider that they describe the "infinite" iteration of a given function. Thus, one considers a sequence $(x_n)$ such that $x_{n+1}=\sqrt{2x_n}$ for every $n\geqslant0$ and $x_0\geqslant0$. Then,  indeed the value of the formula is $2$ in the following sense:

For every $x_0\gt0$, $x_n\to2$.

To see this, note that $x_{n+1}=u(x_n)$, where the function $u:x\mapsto\sqrt{2x}$ is such that $x\lt u(x)\lt2$ for every $x$ in $(0,2)$, $u(2)=2$, and $2\lt u(x)\lt x$ if $x\gt2$. Thus $(x_n)$ is increasing if $0\lt x_0\lt2$ and decreasing if $x_0\gt2$ and converges to $2$ in both cases (and I will let you solve the case $x_0=2$).

Does this work for other numbers?

Indeed, if one means by this the fact that:

$\sqrt{a\sqrt{a\sqrt{a\sqrt{a\sqrt{a\sqrt{\cdots}}}}}}=a$, for every $a\geqslant0$.

To see this, consider $a\gt0$ and $x_{n+1}=u_a(x_n)$, where $u_a$ is the function $u_a:x\mapsto\sqrt{ax}$. Note that $x\lt u_a(x)\lt a$ for every $x$ in $(0,a)$, $u_a(a)=a$, and $a\lt u_a(x)\lt x$ if $x\gt a$. Thus $(x_n)$ is increasing if $0\lt x_0\lt a$ and decreasing if $x_0\gt a$ and converges to $a$ in both cases (and I will let you solve the case $x_0=a$).
A: Let $x=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots2}}}}}}$ then
$x^{2}=2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots2}}}}}}=2x$
so $x(x-2)=0$ since $x\neq 0$ 
$x=2$.
A: $$k=\sqrt{a\sqrt{a\sqrt{a\cdots}}}$$
$$k^2/a=k$$
$$k/a=1$$
$$k=a$$
