Evaluating tetration to infinite heights (e.g., $2^{2^{2^{2^{.^{.^.}}}}}$) The Problem
How can you evaluate (i.e., get a value for) Tetration (i.e., iterated exponentiation) to infinite heights?
For example, what would be the value of this expression?
$$ 2^{2^{2^{2^{2^{.^{.^.}}}}}} $$
My (pathetic) Attempts
I tried equating it to $x$ and substituting it on the RHS, but no luck:
$$ x = 2^{2^{2^{2^{2^{.^{.^.}}}}}} $$
$$ x = 2^x $$
$$ x = \log_{2}{x} $$
What I think we need to do is have the RHS as a polynomial with one variable and the RHS a constant so we can solve for $x$.
I tried drawing the equation on Wolfram Alpha but the lines on the graph don't touch, so no luck there either.
Novice mathematician here. Thanks.
Edit
Sorry, I am a dolt :(
I didn't realize this was a diverging series. What confused me is my math sir told me it could be done. What he actually meant was that it could be said like this:
$$ \frac{\log{x}}{x} = \log{2} $$
but I somehow assumed there would be a numerical answer.
Alternative Questions
@Clayton's answer suggested an similar question which was a convergent series. While that wasn't what my sir meant, it practically could've been:
$$ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}} $$
Another one I can think of would be:
$$ \sqrt{2*\sqrt{2*\sqrt{2*\sqrt{2*\sqrt{...}}}}} $$
Anyway, interesting question this has turned out to be...
 A: Any finite height tower can (in theory) be evaluated.  It will equate to some natural number.  If the tower is only moderately tall, it will be an enormous number.Wolfram Alpha shows that a tower only five layers high has $19729$ digits.  If the tower height is infinite, the value diverges (quickly) to infinity and the value cannot be evaluated.
Your trick of equating to $x$ and substituting will find the limit if it exists.  In this case, it does not.
A: Perhaps your 'math sir' at school meant to tell you $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}$$can be evaluated. In fact, it can be evaluated in the following sense; for $x>0$,  $$x^{x^{x^{.^{.^.}}}}=2\Longrightarrow x^2=2\Longrightarrow x=\sqrt{2}.$$ Hence, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$
A: Hint: Equations $y=x$ and $y=2^x$ do not intersect, means there's no solution for $x\in\mathbb R$.
A: In fairness, WolframAlpha does give you the answer when you type in "x=2^x".  Here's the output I see:

Now, the "Solution" might look a little strange but, if I hit the "Approximate form" I see that it's approximately $0.824679+1.56743 i$ - a complex number.
A: The limit as $n$ approaches infinity of $\, ^nz$ (the $n$ times iterated exponential or power tower function) converges for the bases
${\displaystyle \textstyle (e^{-1})^{e}\leq z\leq e^{e^{-1}}}, \tag*{}$
where
$\displaystyle e^{-e}=\frac{1}{e^e} \approx 0.0659880358453125 \tag*{}$
$\large e^{e^{-1}}=  e^{ \frac{1}{e}} \approx 1.44466786100977 \tag*{}$
It can be verified that when the value of $z$ is between the two numerical values above, the infinite iterated exponential $\, ^{\infty }z$ or $z^{z^{z^{\cdots\infty}}}$ (determined by the relation involving the product log shown below) has a real valued numerical solution.
The infinite power tower or infinite tetration ${\displaystyle z^{z^{z^{\cdot ^{\cdot ^{\cdot }}}}}\!}$ can be extended to complex numbers or to the complex plane.
This infinite power tower has the general value
${\displaystyle h(z)={\frac {W(-\ln(z))}{-\ln(z)}}} \quad (1), \tag*{}$
where $\ln(z)$ is the principal branch of the complex logarithmic function, and $W(u)$ is the Lambert W function or the product log function, defined as:
${\displaystyle u = y e^y \Leftrightarrow y = W(u).} \tag*{}$
For additional clarification, here is the proof of the relation $(1)$ above:
Let
${\displaystyle z^{h}=h} \tag*{}$
Then
${\displaystyle z=h^{1/h}} \tag*{}$
${\displaystyle z^{-1}=h^{-1/h}} \tag*{}$
${\displaystyle {\frac {1}{z}}=\left({\frac {1}{h}}\right)^{1/h}} \tag*{}$
${\displaystyle -\ln(z)=\left({\frac {1}{h}}\right)\ln \left({\frac {1}{h}}\right)} \tag*{}$
${\displaystyle -\ln(z)=e^{\ln \left({\frac {1}{h}}\right)}\ln \left({\frac {1}{h}}\right)} \tag*{}$
${\displaystyle \ln \left({\frac {1}{h}}\right)=W(-\ln(z))} \tag*{}$
${\displaystyle {\frac {1}{h}}=e^{W(-\ln(z))}} \tag*{}$
${\displaystyle {\frac {1}{h}}={\frac {-\ln(z)}{W(-\ln(z))}}} \tag*{}$
${\displaystyle h={\frac {W(-\ln(z))}{-\ln(z)}}.} \tag*{}$
Below is a plot of $(1)$ (from wolfram Alpha):

For $z = 2$ we have the following result:
$\boxed{\displaystyle  {2^{2^{2^{2 ...}}}= h(2) = -\frac{W(-\ln (2))}{\ln (2)}}=\frac{W_0(-\ln (2))}{\ln (2)}}, \tag*{}$
where $W_0$ is the main branch of the Lambert W function.
The numerical value of $h(2)$ is (verified with Mathematica):
$\displaystyle \begin{align} 
 2^{2^{2^{2 ...}}} &= h(2) \\ &\approx 0.8246785461420742223140645943816032399746074201816 \\ &\quad -1.5674321238496478610585743911929869275333075742042 i \end{align} \tag*{}$
Wolfram Alpha providess the following symbolic result :
$\displaystyle -\frac{W(-\ln (2))}{\ln (2)}= \frac{1}{\pi \ln(2)} \int_{-\infty }^{-\frac{1}{e}} \ln \left(\frac{x+\ln (2)}{x}\right) \Im\left(\frac{\partial W(x)}{\partial x}\right) \, dx  ,\tag*{}$
where $\Im(z)$ is the imaginary part of $z$ .
The last result above can be numerically verified and approximated with Mathematica by typing :
NIntegrate[
  Im[D[ProductLog[x], {x}]]*
   Log[(x + Log[2])/x], {x, -Infinity, -(1/E)}]/(Pi*Log[2])

For another value such as $z = \sqrt{2},$ we get:
$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}} = h(\sqrt{2})= -\frac{W(-\ln (\sqrt{2}))}{\ln (\sqrt{2})}=2 \tag*{}$
