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
 A: I think you started correctly by writing that $\exp(z)=\exp(a+ib)=\exp(a)\exp(ib)$ and noting that $\exp(a)$ can never be zero for any $a$. Then you only have to show that $\exp(ib)$ also can never be zero for any $b$ and the proof is complete. This is easily shown using Euler's formula: $\exp(ib)=\cos(b)+i\sin(b)$. In order for $\exp(ib)$ to be zero then both $\cos(b)$ and $\sin(b)$ must be zero for the same $b$, which is not possible for any $b$ since we know that when one of these functions is 0 the other is 1 or -1.
A: Using the given definition,
$$\exp z := \sum_{k = 0}^{\infty} \frac{z^k}{k!},$$
we have
$$\exp z \exp (-z) = \sum_{k = 0}^{\infty} \frac{z^k}{k!} \sum_{l = 0}^{\infty} \frac{(-z)^l}{l!} = \sum_{k = 0}^{\infty} \sum_{l = 0}^{\infty} \frac{(-1)^l z^{ k + l}}{k! l!} .$$
Absolute convergence permits us to reorder the double summation in the second equality above, and also allows us to write it via reindexing as
$$\sum_{p = 0}^{\infty} \sum_{q = 0}^p \frac{(-1)^q z^p}{(p - q)! q!} .$$ Substituting gives that the $p = 0$ term is $1$. For $p > 0$ the coefficient of $z^p$ (in the outer sum) is, by definition of $\cdot \choose \cdot$,
$$\sum_{q = 0}^p \frac{(-1)^q}{(p - q)! q!} = \frac{1}{p!}\sum_{q = 0}^p {p \choose q} 1^{p - q} (-1)^q .$$ We've written the sum on the r.h.s. so that we can recognize it as the binomial expansion for $(1 - 1)^p = 0$, so for $p > 0$ the coefficient of $z^p$ in the series is $0$, hence $$\exp z \exp(-z) = 1 .$$ In particular, there is no $z$ such that $\exp z = 0$.
A: $\exp(z)\neq 0 $ $\space , \forall z\in \Bbb{C}$
Proof:
$f(z) =\exp(z) $ is an entire function i.e holomorphic on $\Bbb{C}$.
To show : $\scr{Z_f}=\{z:f(z)=0\}=\emptyset$
We argue by contradiction. Suppose$\scr{Z_f}\neq\emptyset$.
Let, $z_0\in \scr{Z_f}$
Then, $f^n{(z_0)}=\exp(z_0) =0$ $ \space, \forall n\in \Bbb{N}$
Since, $f\in H(\Bbb{C})$, then $f$ has a Taylor Series  about the point $z_0$ converges for all $z\in \Bbb{C}$.
$f(z) =\sum_{n=0}^{\infty}\frac{f^n(z_0) }{{n!}}(z-z_0)^n$$\space , \forall z\in \Bbb{C}$
Hence, $f(z) =0$ on $\Bbb{C}$.
But, $f(0) =\exp(0) =1$ ,$\space $ a contradiction.
Hence, $\scr{Z_f}=\{z:f(z)=0\}=\emptyset$
In other words, $\exp(z) \neq 0$ for any $z\in \Bbb{C}$
A: If you know that $\exp(z+w)=\exp(z) \exp(w)$, then  $\exp(z)\ne 0$ follows from
$$1=\exp(0)=\exp(z-z)=\exp(z)\exp(-z)$$
A: 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
