Density of $\mathbb{Q}^n$ in $\mathbb{R}^n$ $\mathbb{Q}$ is dense in $\mathbb{R}$ (with the standard topology). I'm pretty sure that $\mathbb{Q}^n$ is dense in $\mathbb{R}^n$ too. Is there an easy argument to prove that without reproducing the same argument to show "$\mathbb{Q}$ dense in $\mathbb{R}$"?
 A: It is. Choose $\epsilon > 0$.   Let $x\in\mathbb{R}^n$.  Then for each k, choose a rational $y_k\in\mathbb{Q}$ so that $|x_k - y_k| < \epsilon/\sqrt{n}.$  Now do the rest.
A: Although the concrete approach given by ncmathadist is certainly an important one, I think the OP is right to (also) ask for a more general argument that "transfers" the density from $\mathbb{Q}$ in $\mathbb{R}$ rather than works again from scratch in the more general case.  Here is a more general statement:

Proposition: Let $\{X_i\}_{i \in I}$ be an indexed family of topological spaces.  If for each $i \in I$ we have a dense subspace $Y_i$ of $X_i$, then $Y = \prod_{i \in I} Y_i$ is dense in $\prod_{i \in I} X_i$ in the product topology.  

Proof: The statement holds trivially when either $I$ or some $X_i$ is empty,* so let's assume all are nonempty.  To show density of a subset of a topological space it is enough to show that it intersects every nonempty element of any given base for the topology (since every nonempty open set contains such a subset).  The standard base for the product topology consists of "cylindrical" sets of the form $U = \prod_{i \in I} U_i$, where $U_i$ is nonempty open in $X_i$ and $U_i = X_i$ for all but finitely many $i$.  Since $Y_i$ is dense in $X_i$, for each $i \in I$ there is a point $x_i \in Y_i \cap U_i$, and then 
$x = (x_i) \in Y \cap U$.  
(In fact the product topology is not really being used here: the same argument shows the stronger statement that $Y$ is dense in $X$ in the box topology.)
Applying this with $I = \{1,\ldots,n\}$, $X_i = \mathbb{R}$ and $Y_i = \mathbb{Q}$ for all $i$ and using that $\mathbb{Q}$ is dense in $\mathbb{R}$, we get an answer to the OP's question. 
*: Can we agree that an empty product of topological spaces should be a one-point space? It seems to me that an empty product in a category should be taken to be the final object of the category, if such a thing exists.  
