Prove that there is a unique analytic function $f: \mathbb{C} \longrightarrow \mathbb{C}$ such that $f'(z) = f(z)$ and $f(0) = 1$.
Hint: Let $g$ be another such function and consider the function $h(z) = f(z)g(-z)$. What do you know about $h(z)$?
My process so far: Let $f(z)=e^z$. I know that $f(z)$ is analytic, and satisfies the above conditions. Taking the hint into consideration I know that $h’(z)=0$ and since $h(z)$ is the product of two analytic functions, its also analytic, implying that $h(z)=k$, and since $f(0)=g(0)=1$ implies that $h(z)=1$. And now I’m stuck. I want to show that this implies $f(z)=g(z)$. Any advise?