$\def\srt{\operatorname{srt}}$
Although there are integral solutions, the super root $\sqrt[n]x_s,\srt_n(x)$ has an expansion using Lagrange reversion:
$$\sqrt[k]x_s=\srt_k(x)= 1+\sum_{n=1}^\infty\frac{\ln^n(x)}{n!}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0$$
Applying general Leibniz rule and $e^z$ Maclaurin expansion each $k-2$ times if $2<k\in\Bbb N$:
$$ \frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^\infty\sum_{n_2=0}^{n-1}\dots\sum_{n_{2k-5}=0}^\infty\sum_{n_{2k-4}=0}^{n-1-\sum_\limits{j=1}^{k-3}n_{2j}}\left.\frac{d^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}{dt^{n-1-\sum_\limits{j=1}^{k-2}n_{2j}}}e^{(n_{2k-5}+1)t}\right|_0\prod_{j=1}^{k-2}\frac{n_{2j-3}^{n_{2j-1}}}{n_{2j-1}!}\left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_{2m}}\\n_{2j}\end{matrix}\right)\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0$$
where $n_{-1}=-n$. Next, use Kronecker delta and factorial power $m^{(n)}$ in $\left.\frac{d^{n_{2j}}t^{n_{2j-1}}}{dt^{n_{2j}}}\right|_0=\delta_{n_{2j},n_{2j-1}}n_{2j-1}^{(n_{2j})}$. As hinted by Quantile Mechanics $(96)$ to $(97)$, remove the odd indexed sums substituting each $n_{2j}=n_{2j-1}$. We reindex $n_{2j}\to n_j$ and simplify:
$$\begin{align}\frac{d^{n-1}\overbrace{e^{t-nte^{te^\cdots}}}^{k-2\ {e^t}\text{s}}}{dt^{n-1}}\bigg|_0=\sum_{n_1=0}^{n-1}\dots \sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j} (n_{k-2}+1)^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-2}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)= \sum_{n_1=0}^{n-1}\dots \sum_{n_{k-1}=0}^{n-1-\sum\limits_{j=1}^{k-2}n_j}(-1)^{n_1}\prod_{j=1}^{k-1}n_{j-1}^{n_j} \left(\begin{matrix}{n-1-\sum_\limits{m=1}^{j-1}n_m}\\n_j\end{matrix}\right)\end{align}$$
where $n_0=n$. These links support the last equality. Therefore:
$$\bbox[1px, border: 2px solid red]{\sqrt[k]z_s=\srt_k(z)=1+\sum_{n=1}^\infty \sum_{n_1=0}^{n-1}\dots\sum_{n_{k-2}=0}^{n-1-\sum\limits_{j=1}^{k-3}n_j}\frac{(-1)^{n_1}\ln^n(z)}{\Gamma \left({n-\sum_\limits{j=1}^{k-2}n_m}\right)(n_{k-2}+1)^{1-n+\sum\limits_{j=1}^{k-2}n_j} n}\prod_{j=1}^{k-2}\frac {n_{j-1}^{n_j}}{n_j!}}$$
where all can be infinite sums or all, except the $n$ sum, can have an upper bound of $n$. Assume improper sums are empty. We find that:
$$\srt_1(z)=z$$
$$\srt_2(z)=1+\ln(z)+\sum_{n=2}^\infty\frac{\ln^n(z)}{n!}(1-n)^{n-1}$$
$$\srt_3(z)=1-\sum_{n=1}^\infty\sum_{k=1}^{n-2}\frac{(-1)^kk^{n-k+1}n^{k-2}\ln^n(z)}{(n-k)!k!}$$
shown here
$$\srt_4(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=1}^{n-k-1}\frac{(-1)^k n^{k-1}k^{m-1}\ln^n(z)}{(n-k-m)!k!m!m^{m+k-n-1}}$$
Shown here
$$\srt_5(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-k-1}\sum_{j=1}^{n-k-m-1}\frac{(-1)^k k^m n^{k-1} m^{j-1}\ln^n(z)}{(n-j-k-m)!m!k!j!j^{j+m+k-n-1}}$$
Shown here
$$\srt_6(z)=1+\sum_{n=1}^\infty\sum_{k=0}^{n-1}\sum_{m=0}^{n-1-k}\sum_{j=0}^{n-1-k-m}\sum_{p=1}^{n-k-m-j}\frac{(-1)^k n^{k-1}k^m m^j j^{p-1}\ln^n(z)}{(n-1-k-m-j-p)!k!m!j!p!p^{p+j+m+k-n-1}}$$
shown here
$$\vdots$$
$$\srt_\infty(z)=\sqrt[z]z$$
Compare both links for verification. The region of convergence is near $|z|=1$. Help is wanted with the region of convergence for the multiple series expansion and other representations of $\srt_k(z)$.
Best Answer
Tetration is a natural extension only of the integer-valued notions of addition, multiplication, etc.
For example, the notion that multiplication is repeated addition fails when you start multiplying by non-integers, and even if you hack together an explanation for the rationals, it requires even more care for the irrationals. The same is also true for exponentiation.
Most things that use these three operations in the physical world involve real-valued operations. For example, your power-law for a fluid viscosity might be $v^{2.4}$, or pressure is related to the inverse of the area, and so forth. Obviously, $e^t$ and its variants appears all the time.
Tetration, however, does not have a unified definition for real or complex heights. As such, its application to our physical domain is quite limited only to areas where things are always integers. And in physics, that happens quite rarely.
Edit: That's not to say that there couldn't in principle be something that is a real number tetrated to an integer height. But -- and there's always a but -- we have to account for units. If you take a length measurement and raise it to the 3rd power, it becomes length-cubed. If you take a length measurement and tetrate it to the 3rd, then it becomes...???