Irreducible Degrees and the Order of a Finite Group  This is a question of aesthetics. 
For a finite group of order $n$, the proof that  the degree $d$ of a complex irreducible representation  divides $n$ goes by showing that the rational number $n/d$ is an algebraic integer. As an application of the fact that  $\mathbf Z$ is an integrally closed domain,  this proof is really spectacular. But I feel that this  is an indirect proof, not providing an insight into what is actually going on.
Being a statement of very basic nature perhaps 
  there are other 'natural'  or alternative ways of seeing why  this happens. I would be grateful if experts here can point out other proofs or explain what is happening in the traditional proof.
 A: Here is an exposition of the ``alternative hack'' to which David Speyer alluded.  For concreteness, I have fixed a group; the argument will go through in general, of course.
Let $g = (12) \in S_3$, a transposition in the symmetric group on three elements.  The conjugacy class of $g$ consists of all three transpositions.  Form $a = \frac{1}{3}\left[(12) + (13) + (23) \right] \in \mathbb{C}S_3$, the average of $g$'s conjugacy class.  It is clear that for any $x \in G$,
$$xax^{-1} = a$$
since conjugating $a$ will simply rearrange the terms in the sum.  Equivalently,
$$xa = ax,$$
and $a$ lies in the center of $\mathbb{C}S_3$.  It follows that if $\rho: S_3 \longrightarrow \mbox{Aut}(V)$ is an irreducible representation, the matrix
$$\rho(a) = \frac{1}{3}\left[\rho(12) + \rho(13) + \rho(23) \right]$$
commutes with any $\rho(x)$.  Schur's lemma tells us that $\rho(a)$ is a scalar matrix, that is, $\rho(a) = \lambda I$ for some $\lambda \in \mathbb{C}$.
Summarizing, for each irreducible representation $\rho$, we may define a class function $\psi_{\rho}$ which associates to any $g \in S_3$ the scalar $\lambda$ by which $a=\frac{1}{|S_3|}\sum_{x \in S_3} xgx^{-1}$ acts on $V$.
As it happens, $\psi_{\rho}(g)$ can be computed easily in terms of the character $\chi^{\rho}$: after all, every element conjugate to $g$ has the same trace; by linearity,
$\mbox{Tr}(\rho(a)) = \chi^{\rho}(a) = \chi^{\rho}(g)$.
It follows that
$$\psi_{\rho}(g) = \frac{\chi^{\rho}(g)}{\mbox{dim}V}.$$
A natural next step is to eliminate reference to a particular $g$ using an inner product:
$$\langle \psi_{\rho},\chi^{\rho} \rangle = \frac{1}{\mbox{dim}V}.$$
Expanding the definition of the inner product on class functions,
$$\mbox{Tr}\left[\frac{1}{|S_3|^2}\sum_{g \in S_3} \left(\sum_{x \in S_3} \rho(xgx^{-1}) \right) \rho(g^{-1}) \right] = \frac{1}{\mbox{dim}V},$$
and
$$\frac{1}{|S_3|^2}\sum_{g \in S_3} \sum_{x \in S_3} \chi^{\rho}(xgx^{-1}g^{-1}) = \frac{1}{\mbox{dim}V}.$$
We are led to consider the formal sum $d=\sum_{g,h \in S_3} ghg^{-1}h^{-1}$ mentioned in David Speyer's post.  This sum is invariant under any automorphism of $S_3$ (in particular inner automorphisms) and so lies in the center of the group algebra $\mathbb{C}S_3$.  Schur's lemma tells us that the image of $d$ under any irreducible representation $\rho$ is a scalar.  By the above we get
$$\sum_{g \in S_3} \sum_{h \in S_3} \rho(ghg^{-1}h^{-1}) = \left[\frac{|S_3|}{\mbox{dim}V}\right]^2I.$$
Considering now the regular representation $\mathbb{C}S_3$, we see that the element $d$ acts with rational number eigenvalues.  In other words, its characteristic polynomial splits completely over $\mathbb{Q}$.  But we can also see by inspection that $d$ acts by an integer matrix.  The characteristic polynomial is a monic integer polynomial and splits into linear factors, so its roots are integers.  Since $\mathbb{C}S_3$ contains every irreducible representation as a summand at least once, we see that each
$$\frac{|S_3|}{\mbox{dim}V} \in \mathbb{Z}.$$
In particular, $1$, $2$, and $1$ all divide $6$.
A: $\def\ZZ{\mathbb{Z}}$I answered a similar question over on math.SE. Some general thoughts:
All proofs I know rely on constructing an element of the group ring which acts by $|G|/\dim V$, or some closely related quantity, on $V$. Since $G$ acts on the regular rep by matrices in $GL_n(\ZZ)$, and $V$ is a summand of the regular rep, this shows that $|G|/\dim V$ is an integer. (Exactly how easy that argument is depends how much algebra your audience has.)
Most books seem to use $\sum_{\chi} \sum_{g \in G} \chi(g) g$, where the sum is over the irreducible characters of $\chi$. This acts by $|G|/\dim V$ on $V$, and clearly lies in $\ZZ^{alg}[G]$, where $\ZZ^{alg}$ is the ring of algebraic integers.
In fact, setting $a_g = \sum_{\chi} \chi(g)$, the $a_g$ are integers, so this element is in $\ZZ[G]$. I spent some time a few months ago trying to find a combinatorial proof that the $a_g$ are integers, without developing the theory of algebraic integers, but failed. For a while I believed that $a_g \geq 0$, but that turned out to be false. Let $X \subset \mathbb{F}_2^5$ be the subgroup $\{ (x_1, x_2, x_3, x_4, x_5) : x_1+x_2+x_3+x_4+x_5=0 \}$ and let $x = A_5 \ltimes X$. If I recall correctly (my notes are at home), I got that $a_g$ is negative at $((12)(34),\ (1,0,0,0,1) )$.
An alternative hack is to use the element $\sum_{g, h \in G} ghg^{-1} h^{-1}$. This is manifestly in $\ZZ[G]$ and it acts by $(|G|/\dim V)^2$ on $V$. That let's you avoid mentioning algebraic integers, but it still involves character theory computations that I find unenlightening.
I'd be interested in hearing other proofs!
