Given that the $n$-th root is defined as follows$$ {\sqrt[n]a} = a^{1/n}$$ Can we infer that n can be irrational, negative or even imaginary? Usually we mean the square root ($n=2$) and then the third root ($n=3$) but is there no reason to think that there could not be $n=2.1$ for example, and that root would be defined as follows:
$$ {\sqrt[2.1]a} = a^{1/2.1}.$$
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Answer has been editted.
First and foremost - notation such as ${\sqrt[2.1]{a}}$, while clunky, is perfectly understandable. There is nothing technically wrong with it. It's just a bit "messy". It's important, however, that we understand what is meant by expressions such as ${a^{\frac{1}{n}}}$, or indeed ${\sqrt[n]{a}}$.
For any positive integer ${n}$, we know that ${a^n}$ is defined as repeated multiplication. For example, ${a^3=a\times a\times a}$. A little thought shows this has the property
$${a^{b}a^{c}=a^{b+c}}$$
(for any integers ${b,c \in \mathbb{N}}$). From this property, we can actually extend the definition of ${a^{n}}$ to also cover powers of the form ${a^{\frac{1}{n}}, n \in \mathbb{N}}$, such as ${a^{\frac{1}{2}}}$. It's any number that satisfies:
$${a^{\frac{1}{2}}a^{\frac{1}{2}}=a^{\frac{1}{2}+\frac{1}{2}}=a^1=a}$$
In the case of ${a^{\frac{1}{3}}}$, it's any number that satisfies
$${a^{\frac{1}{3}}a^{\frac{1}{3}}a^{\frac{1}{3}}=a^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=a^1=a}$$
The problem is, this number need not actually be unique. In the case of ${n=2}$, in fact we can find a concrete example. When we talk about ${4^{\frac{1}{2}}}$, do you mean ${2}$ or ${-2}$? Since both ${2\times 2=4}$, and ${(-2)\times (-2)=4}$. This is where ${\sqrt{a}}$ and ${a^{\frac{1}{2}}}$ differs slightly in typical usage; ${\sqrt{a}}$ by tradition is the positive real value that satisfies the property (also called the "principle value"), and so ${\sqrt{4}=2}$ is uncontroversial, and is accepted. Although ${a^{\frac{1}{2}}}$ can be used to mean "the set of all values that satisfy the property ${a^{\frac{1}{2}}a^{\frac{1}{2}}=a}$", in which we are no longer necessarily talking about a single value.
You may think we are in the clear for ${a^{\frac{1}{3}}}$; since there's only one real value. But of course, we can take things to the complex plane, in which there will be more numbers that satisfy ${a^{\frac{1}{3}}a^{\frac{1}{3}}a^{\frac{1}{3}}=a}$ (two more, in fact).
Ultimately, what notation you accept and use is up to you. But typically, ${\sqrt[n]{a}}$ isn't used to encapsulate all possible answers, only the principle one.
Anyways, we can continue extending. A bit more thought shows for integers ${b,c \in \mathbb{N}}$ we have
$${(a^{b})^{c}=a^{bc}=(a^{c})^{b}}$$
And so further extending, we have for any rational number ${\frac{p}{q}}$ (which ${2.1}$ is, for example), you can simply write
$${a^{\frac{p}{q}}=\left(a^{\frac{1}{q}}\right)^{p}}$$
Again, this doesn't need to be a single value, and could ultimately refer to a set of solutions!
Now, extending for irrational numbers. To calculate ${a^k}$ for some irrational quantity ${k}$, we use rational approximations of $k$ and take the principle values (so we are dealing with positive real numbers). In fact, if we have a series of rational approximations ${p_n}$ such that ${p_n\rightarrow k}$ where ${k}$ is irrational, by the continuity of ${y=a^{x}}$ over the reals we can define
$${a^{k}=\lim_{n\rightarrow \infty}a^{p_n}}$$
(and in fact, you can always define such a sequence. For example, consider ${\pi_n}$ as a sequence of rational approximations for $\pi$. You could define ${\pi_1=\frac{3}{1}}$, ${\pi_2=\frac{31}{10}}$, ${\pi_3=\frac{314}{100}}$... by construction, this will converge to ${\pi}$, but ${\pi_n}$ is rational ${\forall\ n \in \mathbb{N}}$).
For negative numbers, well ${a^{-n}=\frac{1}{a^{n}}}$. Why this? Well, we typically use ${a^{0}=1}$ (unless ${a=0}$, in which case it's undefined - but that's a completely different discussion. You can read more here https://www.math.toronto.edu/mathnet/questionCorner/powerof0.html), and so it makes sense that ${a^{-n}a^{n}=a^{n-n}=a^{0}=1}$ and so ${a^{-n}=\frac{1}{a^n}}$.
For imaginary numbers, this is where papa Euler comes in. Using his formula ${e^{ix}=\cos(x)+i\sin(x)}$, it's possible to define complex exponentiation (I won't go into detail for this here, since this post is already long enough - but I think it's helpful to know how we actually can extend powers to the complex plane).
Of course the notation is valid. However one must be careful that notation is clear, and I fear something like $$\sqrt[\sqrt 2]2$$ is not as clear as $$2^\frac{1}{\sqrt 2}$$
So, while valid, it's rather clunky in my opinion.