Let \(x^{x^{x\cdots}} = 2\). Then \(x^{x^{x\cdots}} = x^{x^{x^{x\cdots}}} = 2\). Substituting, we get \(x^2=2\) and \(x=\sqrt{2}\). Now let \(x^{x^{x\cdots}} = 4\). Similarly, \(x^{x^{x^{x\cdots}}} = 4\), \(x^4=4\) and \(x=\sqrt{2}\). But then \(2=x^{x^{x\cdots}}=4\) and \(2=4\).

Obviously 2 cannot equal 4.

The problem seems to occur with solving for \(x\) by inverting the function \(f(x)=x^{x^{x\cdots}}\) via the recursion \(x^{f(x)}=f(x)\), \(x=f(x)^{1/f(x)}\). While we can set \(f(x)\) to any value \(s\) to solve \(x=s^{1/s}\), we mustn’t neglect also the condition that \(s=f(x)\) should have a solution.

Here we plot \(x=s^{1/s}\) for \(s\) from 0 to 5. (Let us only consider \(x\ge 0\), where everything is real.) Both \(s=2\) and \(s=4\) give \(x=\sqrt{2}\), but only the former satisfies \(s=f(x)\), while no \(s\) to the right of the peak has a solution. The reason is that \(f(x)\) for \(x>1\) has to be increasing by the monotonicity of exponentiation in each of its two arguments (\(a^b\) increasing in both \(a,b\) if \(a,b>1\)), and so the proper inverse (where it is defined) can only be taken from the left branch of the plot. The peak occurs at \(s=e\), \(x=K=e^{1/e}\). For every \(x>K\), \(f(x)\) diverges to \(\infty\). For \(1< x\le K\), \(\overbrace{x^{x^x\cdots}}^n\) is increasing in the tetration depth \(n\) and has an upper-bound (\(e\)), so \(f(x)\) is a well defined limit which is exactly \(s\).

The other half is more complicated. When \(x=0\), \(0^s=0\) for all \(s>0\) and \(0^s=1\) for \(s=0\), so there is no solution to \(s=f(x)\). This is a consequence of where \(a,b<1\), \(a^b\) is increasing in \(a\) while decreasing in \(b\). From \(x<1\), we get \(x^x>x\), \(x^{x^x} < x^x\), and so on in this alternating fashion. We also get from \(x^x < 1\) that \(x^{x^x} > x\), so in fact we have two non-crossing monotonic subsequences — the odd-depth tetrations increasing and the even-depth ones decreasing — that must therefore both converge. \(f(x)\) is well defined iff they converge to the same value \(s\). Suppose one subsequence has limit \(s_1\) and the other \(s_2\). Either of the cases \(s=s_1\) or \(s=s_2\) satisfies \(x^{x^s}=s\), so where there is a unique solution for a given \(x\), the limits must equal and \(f(x)\) must be well defined. From the plot (no proof) this occurs from \((1,1)\) down to the intersection at \(s=1/e\), \(x=L=(1/e)^e\). So for \(L \le x < 1\), \(f(x)\) is well defined. For all \(x < L\) though, multiple solutions for \(s\) actually correspond to the curved branch of this \((s_1,s_2)\) plot, which shows solutions to \(x^{s_1}=s_2\) and \(x^{s_2}=s_1\), i.e., \(s_1^{1/s_2}=s_2^{1/s_1}\), so \(f(x)\) is not defined there. (*)

In summary, \(f(x)\) is well defined between \(1/e\) and \(e\) for \((1/e)^e\le x \le e^{1/e}\), goes to \(\infty\) over the range, and flips between two values under the range. Since \(x^{x^{x\cdots}}=4\) can never happen, it is a vacuous assumption that can logically lead to any conclusion.