We must go deeper

we know x=y^z, y=z^x, z=x^y

take x=y^z

=>x=(z^x)^z

=>x=z^xz

=>x=(x^y)^xz

=>x=x^xyz

=>xyz=1

therefore (xyz)^any number=1

because

1^any number=1

x=y^z => log x = zlogy Similarly, log y = x log z And, log z = y log x

Multiplying above 3 log eqns -> (log x)(log y)(log z) = xyz (logx)(log y)(log z) So, xyz = 1; And hence answer is 1

We claim that $xyz = 1$ .

Suppose some of $x,y,z$ is equal to $1$ ; without loss of generality , suppose $x = 1$ . Then $z = x^y$ implies $z = 1$ , and $y = z^x$ implies that $y = 1$ . This gives $xyz = 1$ .

Otherwise, from $x = y^z$ and $y = z^x$ , we get $x = (z^x)^z = z^{xz}$ . Together with $z = x^y$ , we get $x = (x^y)^{xz} = x^{xyz}$ . Since $x > 0$ and $x \neq 1$ , we can apply logarithm of base $x$ to both sides, giving $\log_x x = \log_x x^{xyz}$ or $1 = xyz$ .

Thus the expression reduces to $1^{1^{1^1}} = \boxed{1}$ .

$y=z^x;\textit{having } z=x^y$ $y=(x^y)^x$ $y= x^{xy};\textit{having } x=y^z$ $y=(y^z)^{xy}$ $y^1=y^{xyz}$ The bases are equal and so the exponents are also equal: $1=xyz$ Having that we can find that the value of the expression is: $1^{1^{1^1}}=\fbox{1}$

(xyz)^(xyz) = x^(xyz) y^(xyz) z^(xyz) =(x^y)^xz (y^z)^xy (z^x)^yz =z^xz x^xy y^yz =(z^x)^z (x^y)^x (y^z)^y =y^z z^x x^y=xyz . So xyz=1.

x^x=1 , lnx^x =ln1 ,x*lnx =0 , lnx=0 , x=1 ( x>0)

The only way for all 3 equations to be satisfied is if x, y, and z are equal to 1. Since 1^1^1^1 is 1, the solution is 1.

Well this is a multiple choice problem, and notice that the first option "1" satisfies the condition if we let x = y = z = 1. Sounds good of a solution to me!