Numbers such that $ \sqrt{ {a_{1}}^{a_{2}^{...^{a_{n}}}}} = a_{1} a_{2} \dots a_{n}$

Question

I wonder whether there are any resources on the equation:

$$\require{\MnSymbol} \require{\mathdots} \sqrt{ {a_{1}}^{a_{2}^{...^{a_{n}}}}} = a_{1} a_{2} \dots a_{n}. \tag{1}$$

Here, $a_{1}, a_{2}, \dots, a_{n} \in \mathbb{N}_{\geq 1}$, and the power tower on the left should have $n$ terms as well -- I wasn't entirely sure how to format it (including the diagonal dots from lower left to the upper right corner) in MathJax. The terms on the right denote a the digits of a single number in base ten.

I found an example of such an identity over here:

$$ \sqrt{ {{{{{{2}^{6}}^{2}}^{1}}^{4}}^{4}} } = 262144. \tag{2} $$

Questions:

1. Are there any resources on equations of the form $(1)$ ?
2. Can you find any solutions, besides the one described in $(2)$? I'm especially interested in solutions that don't involve the number one -- it feels a bit like cheating in the left side of the equation.

Answer 1

Regarding question 2, there are no solutions with every $a_i>1$. The rapid growth of power towers allows for an exhaustive search within some easily bounded parameters. Here's a summary of "near solutions", those with a left side that evaluates to an integer with the correct number of digits:

It's not possible with for $n\geq 5$, since $\sqrt{2^{2^{2^{2^2}}}} > 10^{9864}$.

If $n=4$, the only expression with the correct number of digits is $\sqrt{3^{2^{2^2}}} = 6561$.

If $n=3$, the only expressions with the correct number of digits are $\sqrt{k^{2^2}}$ for $32 \leq k \leq 316$; $\sqrt{4^{3^2}} = 512$; $\sqrt{2^{4^2}} = 256$; $\sqrt{4^{2^3}} = 256$; $\sqrt{5^{2^3}} = 625$; and $\sqrt{2^{2^4}} = 256$.

A solution with $\sqrt{k^{2^2}}$ would satisfy $k^2 = 100k + 22$, which has no integer roots.

If $n=2$, the only expressions with the correct number of digits are $\sqrt{k^3}$ for $k = j^2$ with $j=3$, $5\leq j \leq 21$, or $32 \leq j \leq 46$; $\sqrt{k^4}$ for $4 \leq k \leq 31$; $\sqrt{4^5} = 32$; $\sqrt{3^6} = 27$; $\sqrt{4^6} = 64$; $\sqrt{2^8} = 16$; $\sqrt{3^8} = 81$; $\sqrt{3^{10}} = 243$; $\sqrt{3^{12}} = 729$; $\sqrt{2^{14}} = 128$; $\sqrt{2^{16}} = 256$; and $\sqrt{2^{18}} = 512$.

A solution with $\sqrt{k^3}$ for $k = j^2$ would satisfy $j^3 = 10j^2 + 3$, which has no integer roots.

A solution with $\sqrt{k^4}$ would satisfy $k^2 = 10k + 4$, which has no integer roots.