Why can't I tick all of these options?

Suppose the iterated exponentiation (we'll call it a 'power tower' from now on) converges to a finite value, call it $y$ . We will assume that both $x$ and $y$ are positive; the power tower does not appear (to me) to yield real numbers for any negative values other than $x = -1$ (anyone know whether that is indeed the case?) and at that value it's easy to see that $y = -1$ . So

$\ x^{x^{x^{\cdot^{\cdot^\cdot}}}} = y$

Since both sides are finite, we can use them as exponents of $x$ , i.e.

$\begin{aligned} (x)^{x^{x^{x^{\cdot^{\cdot^\cdot}}}}} &= (x)^y \\ \\ y &= x^y \\ \\ y^{\frac{1}{y}} &= x \end{aligned}$

We derive the left side:

$\begin{aligned} \dfrac{d}{dy} \left( y^{\frac{1}{y}} \right) &= \dfrac{d}{dy} \left( e^{\frac{1}{y} \ln y} \right) \\ \\ &= e^{\frac{1}{y} \ln y} \dfrac{d}{dy} \left( \dfrac{1}{y} \ln y \right) \\ \\ &= y^{\frac{1}{y}} \left( \dfrac{1 - \ln y}{y^2} \right) \end{aligned}$

This yields the critical value $y = e$ . We note that $\lim_{y \to 0^+} y^{\frac{1}{y}} = 0$ ; also, $\lim_{y \to \infty} y^{\frac{1}{y}} = 1$ (see note below). Since $e^{\frac{1}{e}} > 1$ , we have that $y^{\frac{1}{y}}$ has an absolute maximum when $y = e$ , which means that our power tower can only converge for $x \le e^{\frac{1}{e}} \\$

It's not difficult to see that for $0 < x \le 1$ , the power tower is bounded above by $1$ ; for $1 < x \le e^{\frac{1}{e}}$ , it seems intuitively obvious that the power tower is a strictly increasing function, and therefore it will achieve its maximum value at the maximum value of $x$ for which it converges. If we assume it does converge for $x = e^{\frac{1}{e}}$ , say to $L$ , then $(e^{\frac{1}{e}})^L = L$ , and by inspection we would have $L = e$ . Unfortunately we haven't, strictly speaking, proven convergence at $x = e^{\frac{1}{e}}$ ; also, when deriving $y$ it is not immediately obvious that $y'$ is strictly positive for $1 < x \le e^{\frac{1}{e}} \\$

So instead, for $1 < x \le e^{\frac{1}{e}}$ , we define a sequence by $a_1 = x$ and $a_{n + 1} = x^{a_n}$ ; then $y = \lim_{n \to \infty} a_n$ . We can show by induction that $\{a_n\}$ is bounded above by $e$ . We have $a_1 \le e^{\frac{1}{e}} \le e$ ; and if $a_n \le e$ , then $a_{n + 1} = x^{a_n} \le ( e^{\frac{1}{e}} )^e = e$ . Thus our power tower cannot yield values higher than $e$ , so the last three answer choices are not possible; it's not difficult to see that $x = \sqrt{2}$ yields the only possible answer $y = \boxed{2} \\ \\$

Note: to calculate $\lim_{y \to \infty} y^{\frac{1}{y}}$ , we let $z = y^{\frac{1}{y}}$ . Then

$\begin{aligned} \ln z &= \frac{1}{y} \ln y \\ \lim_{y \to \infty} \ln z &= \lim_{y \to \infty} \frac{\ln y}{y} \\ &= \lim_{y \to \infty} \frac{\frac{1}{y}}{1} \qquad \small \text{[Applying L'Hôpital's Rule]} \\ &= 0 \end{aligned}$

Then

$\begin{aligned} \lim_{y \to \infty} y^{\frac{1}{y}} &= \lim_{y \to \infty} z \\ &= \lim_{y \to \infty} e^{\ln z} \\ &= e^0 \\ &= 1 \end{aligned}$

Let $y = x^{x^{x^{\cdot^{\cdot^\cdot}}}}$ . Then $y=x^y$ . We note that $x= \sqrt[y] y$ is a solution, since $y = (\sqrt[y] y)^y$ . The means that $x = 1, \sqrt 2, \sqrt[3] 3, \sqrt[4] 4, \sqrt[5] 5, \cdots$ are likely solutions when the values of $y = 1, 2, 3, 4, 5, \cdots$ respectively. But for $y$ to converge Lambert W-function (see equation 13) requires that $e^{-e} \le x \le e^\frac 1e \approx 0.0659 \le x \le 1.4446$ , and only $x=1, \sqrt 2$ fall into the range and therefore, the other value of $y=\boxed{2}$ .