Answer
$f_n(x)=x^{2^{n+1}}$
Work Step by Step
For $n=0$, $f_1(x)=f_0(f_0(x))=f_0(x^2)=(x^2)^2=x^{2\cdot 2}=x^{2^2}$
For $n=1$, $f_2(x)=f_0(f_1(x))=f_0(x^{2^2})=(x^{2^2})^2=x^{2^2\cdot 2}=x^{2^3}$
For $n=2$, $f_3(x)=f_0(f_2(x))=f_0(x^{2^3})=(x^{2^3})^2=x^{2^3\cdot 2}=x^{2^4}$
Continuing these processes, we find $f_n(x)=x^{2^{n+1}}$ for $n\geq 3$.
Thus, for $n=0,1,2,\ldots$, we get the formula $f_n(x)=x^{2^{n+1}}$.
