Natural derivation of the complex exponential function? Bourbaki shows in  a very natural way that every continuous group isomorphism of the additive reals to the positive multiplicative reals is determined by its value at $1$, and in fact, that every such isomorphism is of the form $f_a(x)=a^x$ for $a>0$ and $a\neq 1$.  We get the standard real exponential (where $a=e$) when we notice that for any $f_a$, $(f_a)'=g(a)f_a$ where $g$ is a continuous group isomorphism from the positive multiplicative reals to the additive reals.  By the intermediate value theorem, there exists some positive real $e$ such that $g(e)=1$ (by our earlier classification of continuous group homomorphisms, we notice that $g$ is in fact the natural log).  
Notice that every deduction above follows from a natural question.  We never need to guess anything to proceed.  
Is there any natural way like the above to derive the complex exponential?  The only way I've seen it derived is as follows:
Derive the real exponential by some method (inverse function to the natural log, which is the integral of $1/t$ on the interval $[1,x)$, Bourbaki's method, or some other derivation), then show that it is analytic with infinite radius of convergence (where it converges uniformly and absolutely), which means that it is equal to its Taylor series at 0, which means that we can, by a general result of complex analysis, extend it to an entire function on the complex plane.  
This derivation doesn't seem natural to me in the same sense as Bourbaki's derivation of the real exponential, since it requires that we notice some analytic properties of the function, instead of relying on its unique algebraic and topological properties.  
Does anyone know of a derivation similar to Bourbaki's for the complex exponential?  
 A: So, what's unnatural about the complex differential equation...
f : C -> C satisfying f'(z) = f(z) and f(0)=1 ?
A: Let

f(x) = cos(x) + i*sin(x)

Then

df/dx = -sin(x) + i*cos(x)
      = i*f(x)

So that

∫(1/f(x)) df = ∫i dx
ln(f(x)) = ix  + C
f(x) = e^(ix + C) = cos(x) + i*sin(x)

Since f(0) = 1, C = 0, so

e^(ix) = cos(x) + i*sin(x)

(I will update this with LaTeX when that functionality becomes available)
A: Also, if you are OK with power series, the Prologue to Walter Rudin's Real and Complex Analysis seems like exactly what you want.  It's a beautiful development of exp, as well as sin, cos, e, and even π, all quite organically.
A: I think essentially the same characterization holds. The complex exponential is the unique Lie group homomorphism from $\mathbb{C}$ to $\mathbb{C}^*$ such that the (real) derivative at the identity is the identity matrix.
A: Some Assumptions
I will assume that you are ok with power series being used, just not Taylor's theorem. I will also assume you will allow us to observe a solution to a DE since you used it in your derivation. 
Defn A series of the form $\sum_{n=0}^{\infty}c_n\left(z-z_0\right)^n$ for $c_n,z,z_0\in\mathbb{C}$ is called a power series.
Thm There is some $R\in[0,\infty]$ such that the power series above converges absolutely for all $z\in\mathbb{C}$ with $\mid z-z_0\mid < R$ and uniformly in $D\left(z_0,\rho\right)$ for all $\rho < R$. Further, the terms are unbounded for all $z$ with $\mid z-z_0\mid > R$.
pf Use the geometric series' convergence.
Lemma Inside the disk of convergence $\sum_{n=1}^{\infty}nc_n\left(z-z_0\right)^{n-1}$ is the derivative of the power series.
Construction
For power functions $y(z)=\sum c_nz^n$ can we find a unique solution to $y'(z)=y(z)$ in $\mathbb{C}$? We can observe that this implies $nc_n=c_{n-1}$. Hence
Defn Let $E\left(z\right):=\sum_{n=0}^{\infty}\frac{1}{n!}z^n$.
Thm 1) $E'=E$
2) $E\left(z_1+z_2\right)=E\left(z_1\right)E\left(z_2\right)$
3)$E_{\mid_{\mathbb{R}}}$ is strictly increasing and $E(\mathbb{R})=(0,\infty)$
4) $x\mapsto E(ix)$ sends $\mathbb{R}$ onto $\mathbb{T}$.
5) There is some real $\pi>0$ such that $E\left(\frac{\pi}{2}i\right)=i$ and for all $z_1, z_2\in\mathbb{C}$, then $E\left(z_1\right)=E\left(z_2\right)$ iff $\frac{z_1-z_2}{2\pi i}\in\mathbb{Z}$
6) $E\left(\mathbb{C}\right)=\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.
Note that all of these can be shown purely at the level of power series using no complex analysis. Further the proofs are not hard, if you want more details here let me know.
Corollary $E$ is a homomorphism of the additive group $\mathbb{C}$ onto the multiplicative group with $\mathbb{C}\setminus\left\lbrace 0\right\rbrace$.
Application
If you care about loops(which I think you do!) lets observe that $\gamma(t)=e^{it}$ for $t\in[0,2\pi]$, is $\mathbb{T}$ (Note here that $\pi$ is simply the real number we found before, not some existential thing!). We could now push this to get winding numbers.
Or you could use this definition of exponential, and the power series definition of inverse to get a (branchless) logarithm. In particular, you can show that the derivative of that fella is $\frac{1}{x}$. So we don't have to define it that way. :)
Comments
I agree with you that the definitions of complex exponentials feel contrived and the logarithm is even worse. The branched logarithms is the only part of Palka that I dislike as a complex book. These definitions I can stomach, as they require no Dues ex Machina.
A: This is more a commentary @Charles Staats answer than a direct answer to the OP. In fact Charles' contribution generated the supplementary question: what about holomorphy? I think we can provide an alternative definition for the complex exponential: 
Definition $\exp$ is the unique holomorphic group homomorphism $\mathbb{C}\to \mathbb{C}^\star$ such that its derivative at the origin is $1$.
It is quite easy to show such a homomorphism is indeed unique. It is less easy to prove its existence if starting from scratch. One way I have found illuminating is the following (based on the Italian classic Analisi matematica by Giovanni Prodi, chapter 6: "Funzioni esponenziali e circolari")
We can show that every continuous group homomorphisms $\mathbb{R}\to \mathbb{C}^\star$ (the latter is the complex multiplicative group) is automatically differentiable (Prodi, §39.1 - sorry, I don't know an English reference. I'm quite sure this is well known, though). Also, it is uniquely determined by its derivative at the origin. This furnishes a (IMHO) very effective definition for real exponential and trigonometric functions: 


*

*$x \mapsto e^x$ is the unique continuous group homomorphism $\mathbb{R}\to \mathbb{C}^\star$ s.t. its derivative at the origin is $1$; 

*$x \mapsto \cos x+i \sin x$ is the unique continuous group homomorphism $\mathbb{R}\to \mathbb{C}^\star$ s.t. its derivative at the origin is $i$.


The images of such homomorphisms are the subgroups $\mathbb{R}^+$ and $\mathbb{S}^1$ respectively. Turns out that their direct product $\mathbb{R}^+\times \mathbb{S}^1$ is the whole of $\mathbb{C}^\star$, and this is why the mapping 
$$f(a+ib)=e^a(\cos b + i \sin b)$$
is a surjective homomorphism $\mathbb{C} \to \mathbb{C}^\star$. We can check directly that it is holomorphic at the origin with complex derivative $1$ and so we can rightfully call it complex exponential.

A substantial difference between this definition and Charles' one is that the adjective holomorphic is really necessary here. To wit, first note that every continuous group homomorphism $f \colon \mathbb{C} \to \mathbb{C}^\star$ is automatically (real) differentiable: if we define two mappings $\mathbb{R} \to \mathbb{C}^\star$ by the equations 
$$f_1(a)=f(a+i0),\ f_2(b)=f(0+ib),$$
we get two continuous group homomorphisms $\mathbb{R}\to \mathbb{C}^\star$ and appealing to the previous differentiability result we easily show that $f_1, f_2$ are $C^\infty$ mappings. And so the same holds for $f$: however, it needs not be holomorphic: take for example
$$f(a+ib)=e^{2a}(\cos b + i \sin b).$$
Ok, I'm finished. Just wanted to share my thoughts with the community. 
