Working from right to left.

[bo198214]

Sorry for putting this in the wrong section BTW.

Im only 14 so I dont understand all this advanced mathematics, but let me ask you this, is x^^0 x?

I put proper parentheses around and apply the law (x^a)^b = x^(a*b):

So, therefore hyper 4 should be:

X^^1=X
X^^2=X^X
X^^3=(X^X)^X = X^(X*X)= X^(X^2)
X^^4=(X^(X^2))^X = X^(X^2*X) = X^(X^3)
X^^5=(X^(X^3))^X = X^(X^3*X)=X^(X^4)

You can continue that ad infinitum and get generally:
x^^n = x^(x^(n-1))
thats what Tetratophile already wrote.

If you apply this to n=0 you get
x^^0 = x^(x^(-1)) = x^(1/x) = \( \sqrt[x]{x} \)

thats exactly your conjecture:

X^^0=X√X

If we put the other negative values into n:

X^^(-1) = X^(X^(-2))=X^((1/x)*(1/x))= \( \sqrt[x]{\sqrt[x]{x}} \)
and so on as you write:

X^^-1=X√X√X
X^^-2=X√X√X√X
X^^-3=X√X√X√X√X
X^^-4=X√X√X√X√X√X
X^^-5=X√X√X√X√X√X√X

but be aware that you must set parentheses otherwise its not clear how to interpret it, e.g. X√X√X can be either X√(X√X) (what is what you meant) but it can also be interpreted as (X√X)√X, whats surely not what you meant.

Note also that the convention among mathematicians is to interpret x^x^x as x^(x^x) instead of what you would suggest (x^x)^x.