jmc

Because $k>0$, $f(x)$ is increasing on the interval $[1,\infty )$. We see that $f(1)=1^k=1$, and as $x$ increases, $f(x)=x^k$ increases without bound. Hence, on the interval $[1,\infty )$, $f(x)$ takes on all values greater than or equal to 1, which means the range of $f(x)$ is $\boxed{[1,\infty )}$.