the exponential is entire and never zero
Two facts about e^z get used so often that they deserve their own entry: e^z is defined and differentiable everywhere, and it is never equal to zero. Both are easy to see once and then become reflexes.
Entire means holomorphic on the whole plane, with no singular points anywhere. The exponential qualifies: from e^z = e^x cos y + i e^x sin y the real and imaginary parts have continuous partials that satisfy the Cauchy-Riemann equations everywhere, so e^z is differentiable everywhere, and a direct computation gives the derivative of e^z equal to e^z again. Never zero is even quicker: |e^z| = e^x, and e^x is a positive real number for every real x, so |e^z| > 0 always. A number whose modulus is positive cannot be zero. Hence e^z omits the single value 0 and hits every other complex number infinitely often (once per horizontal strip of height 2 pi).
These two facts do real work. Because e^z is entire and zero-free, expressions like 1 / e^z = e^(-z) are entire too, and e^z can sit safely in a denominator. And because e^z never hits 0, the logarithm only ever needs to be defined on the punctured plane, the plane with the origin removed; you never have to take log of zero. The fact that an entire non-constant function can omit a value at all is special; Picard's theorem later says it can omit at most one, and 0 is exactly the one value e^z omits.
Solve e^z = 0. Taking modulus, e^x = 0, impossible for real x. So the equation has no solution at all.
There is no complex z with e^z = 0; the value 0 is simply outside the range.
Entire is much stronger than just continuous: it means complex-differentiable on all of the plane, which forces e^z to be infinitely differentiable and equal to its own Taylor series everywhere.