Another (non-homological) proof of the invariance of dimension the only proof I know of that $\mathbb{R}^n$ cannot be homeomorphic to $\mathbb{R}^{n+1}$ requires quite a lot of mathematical apparatus, namely singular homology theory with its exact sequences and pretty abstract setting, as well as some previous (though admittedly not too difficult) calculations of the homology groups of spheres to be used after "excising" a point in $\mathbb{R}^n$.
Is anybody aware of an alternate route? Original research papers are not a problem, as long as they don't involve even more elaborate frameworks or some arcane theory, since eluding those is precisely the point. (To the algebraic-theorists around: it's not that I have any particular bad feelings about homology groups, just curiosity ;-)
Thanks,
 A: You could use homotopy groups instead :).
More seriously, there is a notion of the topological dimension of a space
(to be distinguished from the combinatorial dimension one studies in algebraic geometry). One can show that n-space has dimension n (cf. the book by Hurewicz and Wallman), which proves invariance of dimension. This can be done in an elementary manner, free of any reference to singular homology or homotopy theory.
A: Sperner showed in his doctoral thesis [Sperner 1928] that invariance of open sets, invariance of domain and invariance of dimension can be proved already with elementary combinatorial methods alone.
see also: Elementary proof of topological invariance of dimension using Brouwer's fixed point and invariance of domain theorems?
[Sperner 1928] Sperner, Emanuel: Neuer Beweis für die Invarianz der Dimensionszahl und des Gebietes. In: Abh. Math. Sem. Univ. Hamburg. Band 6, 1928, 265–272
Sperner, Emanuel: Gesammelte Werke. Benz, W.; Karzel, H.; Kreuzer , A. (Hrsg.). Heldermann Verlag, Lemgo, Germany, 2005
A: I don't know if this is what you are looking for, but there is a proof using invariance of domain: if U is  an open  subset of R^n and f:U-->R^n is a continuous injection,  then V=f(U) is open in R^n, and f:U-->V is an embedding. IOW:if U is open in R^n and f: R^n --> R^m is injective and continuous, then f: U --> f(U) is a homeomorphism.
Use the (assumed) existence of this homeomorphism , together with the inclusion map i:R^n-->R^m : (x1,...,xn)-->(x1,..,xn,0,..,0) and then i(R^n) is open in R^m, which cannot happen.
Sorry, I am in a hurry at the moment to give more detail; will come back later.
