TITLE: Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$?
QUESTION [25 upvotes]: This question was prompted by my answer to this question.
An exotic $\mathbb{R}^4$ is a smooth manifold homeomorphic to $\mathbb{R}^4$ which is not diffeomorphic to $\mathbb{R}^4$ with its standard smooth structure. An exotic $\mathbb{R}^4$ is said to be small if it can be embedded in the standard $\mathbb{R}^4$ as an open subset. An exotic $\mathbb{R}^4$ which is not small is called large.

Does every large $\mathbb{R}^4$ embed in $\mathbb{R}^5$?

Freedman and Taylor showed there is a maximal exotic $\mathbb{R}^4$, into which all other $\mathbb{R}^4$'s can be smoothly embedded as open subsets. So it would be enough to show that this one embeds in $\mathbb{R}^5$.
By the Whitney Embedding Theorem, every large $\mathbb{R}^4$ embeds in $\mathbb{R}^8$, but I suspect one can do better.

REPLY [21 votes]: The product of any two smooth open contractible manifolds is diffeomorphic to a Euclidean space, see e.g. remark 5.3 in  my paper; the result is of course not due to me but I don't know any other place where all references are collected. 
In particular, this applies to the product of $\mathbb R$ and any exotic $\mathbb R^4$, so the latter embeds into $\mathbb R^5$. In this case the point is that the product is simply-connected at infinity, and hence by a result of Stallings it is PL homeomorphic to $\mathbb R^5$, but any PL structure on $\mathbb R^5$ is induced by a unique smooth structure (as proved by Munkres).