Shortest/Most elegant proof for $L(1,\chi)\neq 0$
I like the proof by Paul Monsky: 'Simplifying the Proof of Dirichlet's Theorem' American Mathematical Monthly, Vol. 100 (1993), pp. 861-862.
Naturally this does maintain the distinction between real and complex as whatever you do, the complex case always seems to be easier as one would have two vanishing L-functions for the price of one.
I incorporated this argument into my note on a "real-variable" proof of Dirichlet's theorem at http://secamlocal.ex.ac.uk/people/staff/rjchapma/etc/dirichlet.pdf .
There are proofs, notably in Serre's Course in Arithmetic which claim to treat the real and complex case on the same footing. But this is an illusion; it pretends the complex case is as hard as the real case. Serre considers the product $\zeta_m(s)=\prod L(s,\chi)$ where $\chi$ ranges over the modulo $m$ Dirichlet characters. If one of the $L(1,\chi)$ vanishes then $\zeta_m(s)$ is bounded as $s\to 1$ and Serre obtains a contradiction by using Landau's theorem on the abscissa of convergence of a positive Dirichlet series. But all this subtlety is only needed for the case of real $\chi$. In the non-real case, at least two of the $L(1,\chi)$ vanish so that $\zeta_m(s)\to0$ as $s\to1$. But it's elementary that $\zeta_m(s)>1$ for real $s>1$ and the contradiction is immediate, without the need of Landau's subtle result.
Added (25/5/2010) I like the Ingham/Bateman method. It is superficially elegant, but as I said in the comments, it makes the complex case as hard as the real. Again it reduces to using Landau's result or a choice of other trickery. What one should look at is not $\zeta(s)^2L(s,\chi)L(s,\overline\chi)$ but $$G(s)=\zeta(s)^6 L(s,\chi)^4 L(s,\overline\chi)^4 L(s,\chi^2)L(s,\overline\chi^2)$$ (cf the famous proof of nonvanishing of $\zeta$ on $s=1+it$ by Mertens). Unless $\chi$ is real-valued this function will vanish at $s=1$ if $L(1,\chi)=0$. But one shows that $\log G(s)$ is a Dirichlet series with nonnegative coefficients and we get an immediate contradiction without any subtle lemmas. Again it shows that the real case is the hard one. For real $\chi$ then $G(s)=[\zeta(s)L(s,\chi)]^8$ while Ingham/Bateman would have us consider $[\zeta(s)L(s,\chi)]^2$. This leads us to the realization that for real $\chi$ we should look at $\zeta(s)L(s,\chi)$ which is the Dedekind zeta function of a quadratic field. (So if one is minded to prove the nonvanishing by showing that a Dedekind zeta function has a pole, quadratic fields suffice, and one needn't bother with cyclotomic fields).
But we can do more. Let $t$ be real and consider $$G_t(s)= \zeta(s)^6 L(s+it,\chi)^4 L(s-it,\overline\chi)^4 L(s+2it,\chi^2)L(s-2it,\overline\chi^2).$$ Unless both $t=0$ and $\chi$ is real, if $L(1+it,\chi)=0$ one gets a contradiction just as before. So the nonvanishing of any $L(s,\chi)$ on the line $1+it$ is easy except at $1$ for real $\chi$. This special case really does seem to be deeper!
Added (26/5/2010) The argument I outlined with the function $G_t(s)$ is well-known to extend to a proof for a zero-free region of the L-function to the left of the line $1+it$. At least it does when unless $t=0$ and $\chi$ is real-valued. In that case it breaks down and we get the phenomenon of the Siegel zero; the possible zero of $L(s,\chi)$ for $\chi$ real-valued, just to the left of $1$ on the real line. So the extra difficulty of proving $L(1,\chi)\ne0$ for $\chi$ real-valued is liked to the persistent intractability of showing that Siegel zeroes never exist.
My preferred proof is to consider $Z(s) := \prod_{\chi} L(\chi, s)$, where the product is over all characters of $\mathbb{Z}/n$, including the trivial character. It is easy to see that $L(s, \chi)$ is meromorphic near $s=1$, and analytic for $\chi$ nontrivial. So, if any of the $L(1,\chi)$'s were zero, then $Z(s)$ would be analytic near $1$.
But $Z(s)$ is the $\zeta$-function of $\mathbb{Z}[\zeta_n]$. If you know the class number formula for $\mathbb{Z}[\zeta_n]$, then you immediately know that $Z(s)$ has a pole at $s=1$. I think the best approach is to prove this class number formula.
If you want, you can also use the idea of the class number proof but cut some corners. Namely:
Discard primes dividing $n$ from the Euler products. The modified $Z(s)$ is the generating function for ideals $\mathbb{Z}[\zeta_n]$ which are relatively prime to $n$, and that also works. This means you only have to analyze how unrammified primes factor.
We have $Z(s) = \sum_I |I|^{-s}$, where the sum is over all nonzero ideals. We can bound this below by the sum over principal ideals. That already blows up as $s \to 1^{+}$.
You only need to prove that the unit group of $\mathbb{Z}[\zeta_n]$ is discrete, which is the easy part of the Dirichlet unit theorem. If there were fewer units than the unit theorem predicts, the sum would blow up even faster.
I've cut these corners when presenting the idea of the class number proof over tea; but I would probably give the right proof if I were teaching a class.
It's easy to show that one cannot have two real characters $\chi, \chi'$ with $L(1,\chi)=L(1,\chi')=0$, as then $\zeta(s) L(s,\chi) L(s,\chi') L(s,\chi \chi')$ would have a simple zero at s=1 while having non-negative coefficients, which is absurd. This is of course a minor variant of the argument that rules out vanishing for a complex character (and its complex conjugate). The intuition here is that the primes can almost conceivably conspire to be almost totally correlated (or more precisely, anti-correlated) with one character, but not with two; see this blog post of mine.
In some sense, the fact that $L(1,\chi) \neq 0$ with at most one exception is the best one can do using without introducing algebraic number theory methods (class number formula) or moving away from s=1 (in particular, using s=1/2 information); this is discussed in this paper of Granville.
The fact that nobody knows how to make the constants in Siegel's theorem (the one on Siegel zeroes) effective seems to me to be a significant upper bound as to how short or elegant a known proof of $L(1,\chi) \neq 0$ can be; at some point one must perform some maneuvre that is very expensive with regards to effective bounds (e.g. moving one's attention from s=1 to s=1/2, or bounding the class number from below by 1).