Suppose the function $f(x)=x^n,n\ne 0$ is differentiable for all $x$. Then $n$ can be any element of the interval:

Given $f(x)=x^n$.The derivative is $f^{\prime }(x)=n{x}^{n-1}$.For $f(x)$ to be differentiable for all $x$ (including $x=0$), the derivative must exist and be finite at $x=0$.If $n-1<0$ (i.e., $n<1$), then $f^{\prime }(x)=\frac{n}{x^{1-n}}$, which is undefined at $x=0$.Therefore, for $f^{\prime }(x)$ to exist at $x=0$, the exponent $n-1$ must be $\ge 0$.
$$n-1\ge 0⟹n\ge 1$$
Thus, $n$ must be in the interval $[1,\infty )$.