How does one prove that$\frac{\mathrm{d}}{\mathrm{d}z}e^z = e^z$, $\forall z\in\mathbb{C}$?
Attempt 1 $$\frac{\mathrm{d}}{\mathrm{d}z}e^z=\lim_{\Delta z\to 0}\frac{e^{z+\Delta z}-e^z}{\Delta z} = \lim_{\Delta z\to 0}e^z\frac{e^{\Delta z}-1}{\Delta z}$$ I assume that $\frac{e^{\Delta z}-1}{\Delta z}\to1$, but since $\Delta z$ isn't necessarily real I don't know how to deal with the limit.
Attempt 2
I considered using the Cauchy–Riemann equation and it's relation to differentiation to evaluate $\frac{\mathrm{d}}{\mathrm{d}z}e^z$ by computing the partial derivatives. Let $f(z)=e^z$.
$$f_x(z) = \lim_{\Delta x\to0}\frac{e^{x+\Delta x+iy}-e^{x+iy}}{\Delta x} = \lim_{\Delta x\to0}e^{x+iy}\frac{e^{\Delta x}-1}{\Delta x}$$ Since $\Delta x\in\mathbb{R}$ this is essentially just a standard limit from calculus since $e^{x+iy}$ is constant and the fraction is real and tends to $1$. So $f_x(z)=e^z$ and I don't know if it's continuous on all of $\mathbb{C}$ but I'll save that one for later. I cannot assume that $f_y(z)=-if_x(z)$ since I would have to assume that $f$ is entire which I have not proven. So I have to compute $f_y(z)$ like I did with $f_x(z)$. $$f_y(z)=\lim_{\Delta y\to0}\frac{e^{x+i(y+\Delta y)}-e^{x+iy}}{\Delta y} = \lim_{\Delta y\to0}e^{x+iy}\frac{e^{i\Delta y}-1}{\Delta y}$$ Here I run into a problem again because of $e^{i\Delta y}$ and I don't know how to compute that limit.
Eulers formula would possibly make things a bit more simple, it usually does, the reason I didn't want to attempt to apply it is because I have not seen a proof of it that is rigorous that I understand. I have seen the Maclaurin series proof but I am not familiar with complex power series and convergence must be proven. Anyway, this post is getting to be a bit too long for such a simple result, but there you have my attempts. Now please help me. How do I prove this?
Note If you don't use the Euler formula, then you didn't define $e^z$ for complex, which explains why you run into so many trouble. Actually, in some sense, the Euler formula is equivalent to what you want to prove:
Lemma Let $f(z)$ be a function which is equal to $e^x$ on $\mathbb R$, and which satisfies $f(z_1+z_2)=f(z_1)f(z_2)$.
Then CR hold everywhere if and only if $$f(iy)=\cos(y)+ i \sin(y) \forall y \in \mathbb R$$
Idea of Proof: Let $f(iy)=u(y)+iv(y)$. Then $$f(x+iy)=e^x(u(y)+iv(y))$$ And hence, CR are equivalent to $$u=v' \\ v'=-u$$
This system of ODE is easy to solve and the general solution is $$u(y)=C \sin(y) +D \cos (y)\\ v(y)=-C \cos(y) +D \sin (y)$$
Therefore, CR holds if an only if we get the above.
Now, since $1=f(0+i0)=u(0)+iv(0)$ we get $u(0)=1, v(0)=0$ which gives $C=0, D=1$.
Therefore CR hold if and only if $$u(y)= \cos (y)\\ v(y)= \sin (y)$$
QED
If you ever understood the Euler formula, you can look at the following as a proof of Euler formula: The Lemma Above has the following immediate consequence:
Consequence: $f(x+iy)=e^x(\cos(y) + i \sin(y))$ if the only function which is entire, equal with $e^x$ on $\mathbb R$ and satisfies satisfies $f(z_1+z_2)=f(z_1)f(z_2)$.
Moreover, this function has the property $$f'(z)=f(z)$$
P.S. Note that the Euler formula is the definition of the complex exponential. Any proof of the Euler formula is actually a proof that this is the only definition which can work. You can use the following approach if you are concerned about circular logic:
Define $f: \mathbb C \to \mathbb C$ $$f(x+iy)=e^x(\cos(y)+i \sin (y))$$
Then, with definition, you can prove that this $f$ satisfy the following properties $$f \mbox{ is entire } \\ f(0)=1 \\ f'=f \\ f(z_1+z_2)=f(z_1)f(z_2) \\ f(x)=e^x \forall x \in \mathbb R$$
Given these properties, it makes sense to call this function $\exp(z)$.