Exponent of a number is a square root? Say you have a number $x^{\sqrt 2}$.
Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this.
I was thinking something like $$x^{2^{1/2}}$$ and extending on that idea, but I don't know how.
Also, what about for $x^{\sqrt 3}$? Or $x^{\sqrt n}$?

$$\sqrt2= 1.4142135623730951...$$
Let $(a_n):1;1.4;1.41;1.412; 1.4121,...$
Then $$x_1=x^{1}$$
$$x_2=x^{a_2}=x^{1.4}=x^{\frac{14}{10}}=x^{\frac{7}{5}}=\sqrt[5]{x^7}$$
$$...$$
$$x_n=x^{a_n}$$
$$...$$
Then $$x^{\sqrt2}=\lim_{n\rightarrow \infty}x_n$$
