I just watched a video by blackpenredpen where he solved this equation. Here are the steps he took (The process transitions from one line to another. He didn't explicitly use an implies or iff sign, so I'll leave it as such):
$x^{x^{x^{x^{x^{...}}}}} = 2$
$x^2 = 2$
$\left[ \begin{array}{l} x = \sqrt{2} \\ x = -\sqrt{2} \end{array} \right. $
At this point, he just crossed out the $-\sqrt{2}$ option.
Could someone please explain why $-\sqrt{2}$ is not a solution? Thank you!
Short answer
$x^{x^{x^{x^{x^{...}}}}}$ is not defined for $x < 0$.
Explanation
How is $x^{x^{x^{x^{x^{...}}}}}$ defined? It is the limit $\lim_{n\to\infty} a_n$ of the sequence \begin{align*}a_0 & = 1 \\ a_1 & = x \\ a_2 & = x^x \\ a_3 & = x^{x^x} \\ a_4 & = x^{x^{x^x}} \\ & \ldots \end{align*} Starting from $a_2$, these are powers whose exponent is not necessarily an integer. Such kinds of general powers are defined only if their base is positive, which for the above powers means $x > 0$.
(The general power $a^b$ is defined as $\exp(b\ln(a))$, and the expression $\ln(a)$ requires $a > 0$.)
Comment 1
In fact, the power $a^b$ can be defined for $a < 0$ in a meaningful way as long as $b$ is a rational number with an odd denomiator, see the comment of the user "Especially Lime" below. For example, $(-8)^{1/3} = -2$ since $(-2)^3 = -8$. However, this is still not enough for our case.
Comment 2
Actually, the condition $x \geq 0$ is not enough to ensure that $x^{x^{x^{x^\ldots}}}$ is well-defined. The reason is that we did not check if the limit $\lim_{n\to\infty}a_n$ actually exists. You can find this discussion in this answer. The result is that the domain of valid numbers $x$ is $$\frac{1}{e^e} \leq x \leq e^{\frac{1}{e}},$$ which approximates to $$0.066 < x < 1.445.$$ Strictly speaking, this discussion is also missing in the video linked in the question: To show that $x = \sqrt{2}$ is a solution of $x^{x^{x^{x^\ldots}}} = 2$, one has to ensure that the limit exists in the case $x = \sqrt{2}$. (Disclaimer: Since the calculation in the video doesn't use any implication signs, we don't know what he claims to have actually achieved.)
Comment 3
The discussion of the domain of $x$ leads to the range of $$x^{x^{x^{x^\ldots}}} = y$$ as $$\frac{1}{e} \leq y \leq e.$$ I find it quite surprising that the range has a finite maximum.
Comment 4
Within the above range of $y$, we can solve the equation $$x^{x^{x^{x^\ldots}}} = y$$ in general, in the same style as as it is done in the video linked in the question. We have $$x^y =x^{\left(x^{x^{x^\ldots}}\right)} = x^{x^{x^{x^\ldots}}} = y$$ and therefore $$x = y^{\frac{1}{y}}.$$ In the case $y = 2$ this reproduces the solution of the original problem as $x = 2^{\frac{1}{2}} = \sqrt{2}$.