jmc

We are looking at the range of $f(x)$ when $x$ is in the interval $[1,\infty )$. Because $k<0$, $f(x)$ is decreasing on the interval $[1,\infty )$. We see that $f(1)=1^k=1$, and as $x$ increases, $f(x)=x^k$ approaches 0, but never reaches it. Hence, on the interval $[1,\infty )$, $f(x)$ takes on all values between 0 (exclusive) and 1 inclusive, which means the range of $f(x)$ is $\boxed{(0,1]}$.