Prove that the exponential $\exp z$ is not zero for any $z \in \Bbb C$ How can the following been proved?

$$
\exp(z)\neq0, z\in\mathbb{C}
$$

I tried it a few times, but I failed. Probably it is extremly simple.
If a draw the unit circle and then a complex number $\exp(a+ib)=\exp(a)\exp(ib)$ then it is obvious that this expression is only $0$ if $\exp(a)$ equals zero, but $\exp(a),a\in\mathbb{R}$ is never zero. This seems not to be very robust.
Thank you

Let’s prove that  ez=0  has no solution  ∀z∈C
We are going to use the fact that if  f:C⟶C  is holomorphic on the whole  C  plane, then it is equal to its taylor series on the whole  C.
First, let’s prove that     f(z)=ez  is holomorphic on the whole     C  using the Cauchy-Riemann Equations, which state that a complex function     f(z)=u(x,y)+v(x,y)i  is holomorphic if and only if the following conditions are satisfied:
∂u∂x=∂v∂y
∂u∂y=−∂v∂x
To write our function  f(z)=ez  separating the real and imaginary parts, we use Euler’s formula:
f(z)=ez=ex+iy=ex(cos(y)+isin(y))=excos(y)+iexsin(y)
Using Cauchy-Riemann Equations:
∂u∂x=excos(y)=∂v∂y
and
∂u∂y=−exsin(y)=−∂v∂x
Since Cauchy-Riemann Equations are satisfied and the partial derivatives are continuous, we can conclude that the function is holomorphic  ∀z∈C
Now, how could we proceed? Well, as I said if a function is holomorphic on the whole complex plane, than all the functional values on the plane are equal to the function’s taylor series. Now, let’s assume:
∃z0∋ez0=0
Then, since all the derivatives of     ez  are itself, all the derivatives are zero at     z0 , which implies that the Taylor series of the function centered at  z0  is identically 0 in the whole complex plane, which is clearly a contradiction. Hence, there doesn’t exist any     z∈C∋ez=0
