Does 2 to the power of root 2 mean anything?
It doesn't make any sense to me when thinking about basic indices?
I've written this before but can't find it now so I will repeat.
I presume that you have been introduced to the power notation as \(\displaystyle x^2= x\cdot x\), \(\displaystyle x^3= x\cdot x\cdot x\), etc. with "\(\displaystyle x^n\) mean "x multiplied by itself n times."
Obviously that requires that n be a positive integer. It just doesn't make sense to "multiply x by itself -1 times" or (your case) "multiply x by itself \(\displaystyle \sqrt{2}\) times" or even "multiply x by itself 0 times".
But there are many times we would like to use them. And, hey, this is mathematics- everything depends on definitions. We are free to define things as we want. As long as everyone agrees we are free to define "\(\displaystyle x^{\sqrt{2}}\)" as we like.
Now, \(\displaystyle x^n\), with n a positive integer, has two nice properties. One is that \(\displaystyle (x^n)(x^m)= x^{n+m}\) because \(\displaystyle x^n\) has n copies of x multiplied, \(\displaystyle x^m\) has m copies, so \(\displaystyle (x^n)(x^m)\) has n+ m copies of x multiplied. The other is that \(\displaystyle (x^m)^n= x^{mn}\). To see that, imagine \(\displaystyle x^m\) as a row of m x's. Taking that to the m power, we are multiplying m of those rows- think of a rectangle with m rows, each with n x's so there are mn x's.
As I said, we are free to define "\(\displaystyle x^r\)" as we please but it would be nice to do it so that those rules, \(\displaystyle (x^n)(x^m)= x^{n+ m}\), and \(\displaystyle (x^m)^n= x^{mn}\) were still true.
"0" is of course the "additive identity", n+ 0= n for any n. In particular \(\displaystyle x^nx^0=x^{n+ 0}= x^n\) so we want \(\displaystyle x^0= 1\) and that is how we define \(\displaystyle x^0\) for any x.
Every positive integer has a "negative" such that \(\displaystyle n+ (-n)= 0\) so we can write \(\displaystyle x^{n}x^{-n}= x^{n- n}= x^0= 1\) so we want \(\displaystyle x^{-n}\) to be the "multiplicative inverse" of \(\displaystyle x^n\) which is \(\displaystyle \frac{1}{x^n}\).
That takes care of the integers. The next set of numbers is the rational numbers. A rational number is a number of the form \(\displaystyle \frac{m}{n}\) where m and n are integers. Now we want to define x to a fractional power so that \(\displaystyle (x^m)^n= x^{m+ n}\). In particular, we want \(\displaystyle (x^{\frac{1}{n}})^n= x^{\frac{n}{n}}= x^1= x\). That means we want \(\displaystyle x^{1/n} = \sqrt[n]{x} \), the nth root of x. Then, of course, \(\displaystyle x^{\frac{m}{n}}= \left(x^{\frac{1}{n}}\right)^m= \left( \sqrt[n]{x} \right)^m= \sqrt[n]{x^m} \)
The irrational numbers, like your \(\displaystyle \sqrt{2}\), cannot be defined "algebraically" as above- they have to be defined "analytically" with some kind of limiting process. The simplest is that every irrational number can be defined as the limit of a sequence of rational numbers. That is what we mean when we say that every irrational number iis For example, "\(\displaystyle \pi\)" is "3.141596..."- it is the limit of the sequence of rational numbers, 3, 3.1, 3.14, 3.141, 3.11415, etc. \(\displaystyle \sqrt{2}= 1.4142...\). That is, \(\displaystyle \sqrt{2}\) is the limit of the sequence of rational numbers 1, 1.4, 1.41, 1.414, 1.4142.
And we define x to an irrational power to make \(\displaystyle x^y \), for variable y, a continuous function. We define \(\displaystyle x^{\sqrt{2}}\) as the limit of the sequence \(\displaystyle x^1\), \(\displaystyle x^{1.4}\), \(\displaystyle x^{1.41}\), etc.
Now, you can't actually calculate the limit of that sequence (and neither can your calculator) but you can approximate it by taking an approximation to \(\displaystyle \sqrt{2}\). My calculator says that \(\displaystyle \sqrt{2}\) is equal to 1.4142135623730950488016887242097. It isn't, of course, but that is the rational number approximation my calculator is using. And \(\displaystyle 2^\sqrt{2}\) is given as 2.6651441426902251886502972498731. Again, that is not the actual value of \(\displaystyle 2^\sqrt{2}\) but is the best approximation it can give.