The answers so far give interpretations of the exponential as a limit of discrete approximations. An alternative interpretation is that any (sufficiently nice) continuous map that takes addition to multiplication on the complex line and takes reals to reals has a purely imaginary kernel isomorphic to the integers. Given The constant $e$ arises from a suitable normalization , for which unit speed paths on the imaginary axis are taken to unit speed paths on the unit circle, so and $\pi$ then shows up as a path length. One way to emphasize the additive-multiplicative relationship is to expand the formula as: $e^{\pi i-0i} = -1/1$.
Here is a more formal treatment: We consider Both $(\mathbb{C}^\times, \times)$ and $(\mathbb{C}, +)$ are one dimensional analytic groups, and the latter is simply connected, so there is a universal covering homomorphism $\exp: \mathbb{C} \to \mathbb{C}^\times$ from the additive group to the the multiplicative group. It The homomorphism is unique once we choose a normalization, e.g., by demanding that it is analytic and satisfies the differential equation $\partial_z \exp = \exp$. (The differential equation can be related to the homomorphism, after choosing coordinates, by considering the respective formal group laws, or just reasoning heuristically with infinitesimals.)infinitesimals.
Remarks: One can also show this by more direct methods, such as applying Note that the above differential equation to grind out only part of the identity normalization we used here was the fact that $\frac{\partial}{\partial y} | \partial_z \exp(iy) |^2 = 0$ (although one should note that the normalization exp$ is not used here). a real multiple of $\exp$. The "defined over the reals" bit may be unsatisfying to some, but you can check the conjugation behavior can be verified directly by expanding as a power series that converges everywhere, noting and checking that the coefficients are real. One can also prove the claim by more direct methods, and checking that it converges everywheresuch as applying the above differential equation to grind out the identity $\frac{\partial}{\partial y} | \exp(iy) |^2 = 0$.
Proof: Using By combining the previous claim concerning unit norms , with the above differential equation implies $\partial_z \exp = \exp$, we conclude that $\exp$ takes any unit speed path on the imaginary axis to a unit speed path on the unit circle. We have $\exp(0) = 1$ by the homomorphism assumption, and the length of a minimal path from $1$ to $-1$ on the unit circle is $\pi$.
Remark
Remarks: Depending on how $z \mapsto e^z$ is defined, one may still have to check that it agrees with $\exp$, but this isn't a big deal. I tried to avoid choosing square roots of minus one as much as possible, but the statement of the identity makes it a bit difficult to maintain such discipline.
