Sufficiently large implies splitting

Statement

Let $G$ be a finite group, and let $d$ be the exponent of $G$: in other words, $d$ is the least common multiple of the orders of all elements of $G$. Suppose $k$ is a sufficiently large field for $G$: $k$ is a field whose characteristic does not divide the order of $G$, and such that the polynomial $x^d-1$ splits completely over $k$.

Then, $k$ is a splitting field for $G$: Every linear representation of $G$ that can be realized over an algebraic extension of $k$ can in fact be realized over $k$.