We assume our base field $k$ is algebraically closed, so any $k$-torus is split. As in the heading above, for $n = 1, \mathbb{G}_m$ is a rational variety and so its function field is $k(t)$ for an indeterminate $t$. Is the function field of $\mathbb{G}_m^n$ simply $k(t_1,...,t_n)$? In other words, are $n$-dimensional tori $k$-rational varieties?
When $k$ is not algebraically closed, this is not true. But it still holds for $n=1$, since up to isomorphism we only have two $1$-dimensional tori. They are the trivial torus and the one given by Weil restriction of a two-dimensional extension of the base field.