An infinitely powered expression Here's an expression I am struggling to evaluate:
$$\LARGE
  {\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\:\cdot^{\:\cdot^{\:\cdot}}}}}}
$$
The value turns out be $2$, but I don't understand how do we get it. Can anyone give the solution?
EDIT: The original problem is as follows: 
If $y(x)= { x }^{ { x }^{ { x }^{ { x }^{.  } } } }$, then evaluate $y(\sqrt { 2 })$.
 A: Really you wish to solve for $x$ in the expression below $$x^{x^{x^x…}}=2, $$ where $x$ exponentiates an infinite amount. One way to approach it would be to note $$\log_x( x^{x^{x^x…}})=\log_x(2)\rightarrow x^{x^{x^x…}}=\log_x(2). $$ But we know $x^{x^{x^x…}}=2 $, therefore, $$2=\log_x(2)\rightarrow x^2=2$$ and thus $$\boxed{x=\sqrt{2}}. $$
This is just a good way to show it. It is certainly not a rigorous proof because you could really perform the logarithm step as many times you like and get different solutions. But, it helps in basic understanding, in my opinion. 
A: Consider the sequence $\{0,1,\sqrt{2},\sqrt{2}^{\sqrt{2}},\sqrt{2}^{\sqrt{2}^{\sqrt{2}}},\ldots\}$. In other words the sequence with $a_0=0$ and $a_{n+1}=\sqrt{2}^{a_{n}}$.
Note that $a_0<2$. Assume that for some $n$, that $a_n<2$. Then $a_{n+1}=\sqrt{2}^{a_n}<\sqrt{2}^2=2$. So by induction, $a_n<2$ for all $n$.
Note that $a_0<a_1$. Suppose that for some $n$, that $a_{n-1}<a_n$. Then $\sqrt{2}^{a_{n-1}}<\sqrt{2}^{a_n}$, so $a_n<a_{n+1}$. So by induction, $a_n<a_{n+1}$ for all values of $n$. In other words, the sequence is increasing.
So we have that this sequence is increasing and bounded above. Therefore it converges to some limit $L$. That means
$$\begin{align}
a_{n+1}&=\sqrt{2}^{a_n}\\
\lim_{n\to\infty}a_{n+1}&=\lim_{n\to\infty}\sqrt{2}^{a_n}\\
L&=\sqrt{2}^L\\
\end{align}$$
By concavity, there are only two solutions to this equation, and fortunately it's easy to identify both: $L=2$ or $L=4$. But $L$ cannot be $4$, since the sequence was bounded by $2$. 
So this establishes that $\lim_{n\to\infty}a_{n}$ exists and that its value is $2$.
A: Let $x=\sqrt{2}^{\sqrt{2}^{\cdots}}$ then $x=\sqrt{2}^x$ or $x=2^{x/2}$ i.e., $x^2=2^x$
A: If we have an infinite number of powers nested then we can say
$$v=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}=\sqrt{2}\,^v$$
Take the log to base 2.
$$lg(v)=v\,lg(\sqrt{2})$$
and
$$\frac{lg(v)}{v}=lg(\sqrt{2})=\frac{1}{2}lg(2)$$
And this gets us $v=2$
