Nonsense, this equation doesn't even make any sense without a well defined notion of the exponential function, and then a well defined extension of said function into the complex numbers. You will already have e by the time you reach this equation because defining what exp(z) even means requires you already know the properties of e^x over the reals you wish to preserve. And Euler's formula comes from finding such a function and then defining it to be exp(z). Multiplication in the complex plane by a unit vector is a rotation. Exponentials "repeated multiplication" by such a vector is spinning. And it turns out spinning at a constant rate satisfies the properties of the exponential function so it makes sense to say that's what exp(ix) means.
This is perhaps the most unnatural equation (well identity) in maths. It doesn't fall out anywhere, you would never write it down and solve for e, it's a special case of a more general result you would get first, and it's symbol soup for precisely the reason that the identity itself confers no understanding.
exp/log are natural because you almost can't help but discover them as they appear in so many different seemingly unrelated places.
> And Euler's formula comes from finding such a function and then defining it to be exp(z).
That's certainly one way, but you can also define exp via its power series (which is easily proven to be convergent everywhere). Then, all the properties of exp, as well as Euler's formula, are actual theorems, not just definitions.
It doesn’t matter which definition you pick. I e.g. had it defined axiomatically when I studied, i.e. we were given a list of properties which identify exp unambiguously and then we were given a proof of its existence. The fact that those properties were part of the definition doesn’t take away from their profoundness. The function could still just not exist. Same thing with defining by formulas. The map is not the territory.
Not really. Your equation has no unknowns, but if you consider the "e" in your equation as the unknown, and solve it (here I use "a" instead as the unknown, and use "e" in its usual meaning):
We want a^(i pi) + 1 = 0. Now,
a^(i pi) = e^(ln(a) i pi) = e^(i ln(a) pi) = cos( ln(a) pi) + i sin( ln(a) pi),
so we want cos( ln(a) pi) = -1, sin( ln(a) pi) = 0,
so ln(a) = 1, 3, 5, so a = e, e^3, e^5, ...
Thus indeed e^(i pi) + 1 = 0.
But also ln(a) = -1, -3, -5 work, so for example for
a = 1/e = 0.3678794412... > 0, we have
a^(i pi) + 1 = 0;
and of course for a = e^-99 = 1.0112214926104485... × 10^-43 etc.