Personally, my intuition for the complex exponential $x\mapsto e^{ix}$ is Lie-theoretic.

A smooth manifold is a topological space which is locally diffeomorphic to Euclidean space; a group is a set with an associative binary operation, inverses and an identity (often we speak of it encoding some form of symmetry; the collection of transformations which preserve a set of properties of a given object, if closed under multiplication and inverses, is a group); a space that is both a smooth manifold and a group in which multiplication and taking inverses are smooth maps is called a Lie group; the tangent space of the identity element (which will become equipped with a 'bracket' operation, but this operation is trivial if our group is abelian) is called a Lie algebra.

The circle group, as a subspace of the complex plane, is clearly a one-dimensional Lie group, and its Lie algebra is therefore a line, so in particular $\cong\bf R$. The preceding paragraph invokes a lot of heavy language to deal with what is essentially a very simple object; placing this example in perspective is something I find helpful though. For any tangent vector $X$ in the Lie algebra / tangent space, there is a unique one-parameter subgroup in the Lie group which, as a curve, has the given tangent vector at the identity. The exponential $\exp(tX)$ is defined to parametrize this curve.

Suppose our Lie group is realized concretely as a matrix group (see e.g. Ado's theorem). Then the tangent space is also a subspace of matrices, and the exponential map satisfies $\Phi'(t)=X\Phi(t)$, where $\Phi(t)=\exp(tX)$, which can be deduced using $\Phi(t+h)=\Phi(t)\Phi(h)$ and $\Phi'(0)=X$. Some calculus tells us we have $\exp(tX)=\sum_{n\ge0}(tX)^n/n!$. Then, using complex numbers and their multiplication as an acceptable substitute for our $2\times2$ matrices, the tangent space of $1$ on the unit circle is identified with the imaginary axis $i\bf R$, and our exponential is in fact $\exp:x\mapsto e^{ix}$.

The mystery, then, lies in how a geometrically-inspired "exponential map" manifests as a number being raised to powers (i.e. a bona fide exponential); this is not so mysterious, though, when we consider the fact that our $\exp$ (when restricted to linear subspaces of the Lie algebra; we actually don't have $\exp(AB)=\exp(A)\exp(B)$ in general - see the BCH formula) is an additive group homomorphism, exactly as a bona fide exponential.