Tetration of Alephs Does this even mean anything?
$\underbrace{x^{x^{x^{...^{x^x}}}}}_n$
Where $n = \aleph_0$?
Because I "know" it converges when (say) $x = .5$ and $n \to \infty$
 A: This is meaningful for ordinal numbers only.  This operation has no meaning in cardinal arithmetic (explained below).  We can define an operation $\uparrow \uparrow$ using transfinite recursion as follows:
$$\alpha \uparrow \uparrow 0 = 1$$
$$\alpha \uparrow \uparrow \beta + 1 = \alpha^{ \alpha \uparrow \uparrow \beta} $$
$$\alpha \uparrow \uparrow \gamma = \sup \{ \alpha \uparrow \uparrow \beta : \beta < \gamma \} \text{ where $\gamma$ is a limit ordinal}$$
EDIT: This is only meaningful if we are talking about this as an operation with ordinal numbers.  This has no meaning for cardinal numbers.  However, we can talk about this operation with the ordinal number $\omega$, for example.
The problem is that we want our hypothetical operation $2\uparrow \uparrow \omega$ or $2 \uparrow \uparrow \aleph_0$ to be a fixed point so that $2^{ 2 \uparrow \uparrow \omega } = 2 \uparrow \uparrow \omega$.  But if $\aleph_\kappa$ is an infinite cardinal, then $2^{\aleph_\kappa} \ne \aleph_\kappa$ so our operation would have no fixed points.
A: I don't know about cardinal numbers, but hyperoperations on ordinal numbers  have been studied, e.g. by John Doner and Alfred Tarski, An extended arithmetic of ordinal numbers, Fund. Math. 65 (1969), 95–127. That paper is freely available here:
http://yadda.icm.edu.pl/mathbwn/element/bwmeta1.element.bwnjournal-article-fmv65i1p10bwm
A: For a cardinal/combinatorial perspective, note that $2 \uparrow \uparrow n = | V_{n+1} | = 2^{|V_n|}$ for finite $n$, where $V_n$ is stage $n$ of the cumulative hierarchy of sets. We could extend this to transfinite $n$. There might also be a sensible way to extend it to bases other than 2.
A problem here is that the arithmetic we would expect seems to be inconsistent for transfinite cardinals. For instance,
\begin{align}
2 \uparrow \uparrow \aleph_0
&= 2 \uparrow \uparrow (\aleph_0 + 1) \\
&= 2^{2 \uparrow \uparrow \aleph_0} \\
&> 2 \uparrow \uparrow \aleph_0 & \text{(Cantor's theorem)}
\end{align}
A: Since $2^{\aleph_0} = \aleph_0^{\aleph_0}$, we can generalize and have that $\wp(κ) = κ ↑↑ 2$. Technically, $\aleph_0 ↑↑ \aleph_0 = \beth_{\omega}$ (I think someone else on this site noted this elsewhere, but I can't find the posting).
You can also use Wolfram Alpha to iterate self-exponentiation of $\aleph_0$ and get the $\beth_n$ from this.
Tetration also shows up in accounts of ultrafilters, via the expression "22κ." See e.g. page 4 of this article.
As for 2 ↑↑ $\aleph_0$, this would depend on what $2^{\aleph_0}$ (as 2 × 2 × ...) is (note that 2 + 2 + ... as 2 × $\aleph_0$ = $\aleph_0$). Presumably, it would be some $\kappa$ greater than $\aleph_0$, but how much so? We wouldn't know...
