Yes, this follows from the homological characterization of regularity. Let $A=k[x_1,\dots,x_n]$ and $B=K[x_1,\dots,x_n]$, where $K$ is the algebraic closure of $k$. Then for any $A$-modules $M$ and $N$ with $M$ finitely generated, $$B\otimes \operatorname{Ext}_A^*(M,N)=\operatorname{Ext}_B^*(B\otimes M,B\otimes N).$$ Since $B$ has global dimension $n$, the right hand side vanishes for $*>n$, and hence so does $\operatorname{Ext}_A^*(M,N)$. It follows that $A$ has global dimension $n$ and is thus regular. Alternatively, there are various ways to prove directly that $A$ has finite global dimension, without reducing to the algebraically closed case (for instance, you can show that the global dimension of $R[x]$ is always one more than the global dimension of $R$).

in finitely many indeterminates... $\endgroup$ – Fred Rohrer Nov 22 '14 at 12:45