Of course, if one were setting out the conventions of complex exponentiation anew, one could make e^(i*x) equal whatever one wanted it to equal. The quite natural condition which leads us to set it out as we do is our desire to preserve the property that the function e^x should be its own derivative and take the value 1 on the input 0, even when considered in the complex context.
It’s essentially the same as some of the proofs linked to before, but just to phrase it in (what I feel to be) a particularly simple manner, I like the following way of proving e^(ix) = cos(x) + isin(x), without using Taylor series and using from the theory of differential equations only the knowledge that certain kinds of systems of equations have unique solutions:
First, we decompose e^(ix), as a complex-valued function, into its real and imaginary components, setting it equal to a(x) + ib(x) for some real-valued functions a and b of x. Since e^0 = 1, it follows that a(0) = 1 and b(0) = 0.
Next, we obtain the derivative of e^(ix). Since e^x has as derivative e^x, it follows that e^(ix) has as derivative ie^(ix), which is therefore equal to i*(a(x) + ib(x)) = -b(x) + ia(x). But since this is the derivative of a(x) + i*b(x), it must be the case that the derivative of a(x) is -b(x) and the derivative of b(x) is a(x).
So this brings us to a pair of differential equations of a certain form relating a and b, along with their “initial values”; by the theory of differential equation, these will determine a unique solution [essentially because, starting at the initial values, one could straightforwardly deterministically “plot out” the rest of a solution, using the information from the differential equations]. And, as it happens, taking a(x) as cos(x) and b(x) as sin(x), we see that they do in fact clearly satisfy the differential equations and the initial values. Thus, we can conclude, e^(ix) = cos(x) + isin(x). Q.E.D.
As for why e^x should be its own derivative and take value 1 on the input 0, well… that’s, let us say, essentially the definition of e^x. “But why should there exist any or a unique e having that property for exponentiation with it as a base?”, you may ask. Well, let us restrict ourselves to the real context for a moment. By essentially the same reasons as invoked above in square brackets, there is a unique function exp from reals to reals satisfying exp(0) = 1 and being its own derivative. We can essentially define e as exp(1). Now, what remains to be shown is that exp(x) = e^x = exp(1)^x. As it happens, a function f(x) is everywhere equal to f(1)^x if and only if f is continuous, f(0) = 1, and f(c+x) = f©f(x) [the latter two conditions inductively show that f produces the value f(1)^x on all naturals x, which straightforwardly inductively extends to this holding for all integers x, and then for all rationals x; the continuity criterion takes it from there to all reals]. exp is continuous because it is, a fortiori, differentiable; furthermore, exp(0) = 1 because that was one of the constraints which defined it. Finally, exp satisfies the last condition because exp(c+x)/exp© is also, by straightforward differentiation rules, its own derivative, and clearly takes value 1 on the input 0; thus, exp(c+x)/exp© = exp(x), giving us the third condition [we can word this slightly differently to get around the worry that exp© might be 0, throwing the division into undefined territory, but, at any rate, this cleanly expresses the idea]. We can conclude, at least in the real context, that exp(x) does indeed equal exp(1)^x, which means, having defined e as exp(1) and combining this with what was shown above, that there is a unique value e such that e^x is its own derivative. Extending this to the complex context is then just a matter of obtaining an appropriate definition of exponentiation with a complex power that maintains the necessary properties, and what was done above in “proving” e^(ix) = cos(x) + isin(x) is essentially to demonstrate that the appropriate general definition of exponentiation is the one essentially given by that equation. [That is, where b^(c + id) = cos(ln(b)c) + isin(ln(b)d). This works quite well if one is just thinking about b as a positive real; more generally, one must confront the question of what the natural logarithm is to mean in more general contexts, which will generally have to involve some notion of a multi-valued function or branch cuts or so on (because, for example, e^0 and e^(2pi*i) take on the same value)].