Starting from the assumption that $e^x$ is the solution to the equations $f(x)=f'(x)$ and $f(0)=1$, how may one derive the direct expansions of $e$?

Question

If it is assumed that $e^x$ is the unique solution to the differential equation $f(x)=f'(x)$, how may we derive from the initial assumption the infinite sum expansion and the infinite product expansion of $e$?

Answer 1

$$f(x)=f'(x)\Rightarrow f'(x)=f''(x)\Rightarrow f''(x)=f'''(x)$$ It follows easily that for all $n$ $$f(x)=f^{(n)}(x)$$ Apply now Taylor expansion.

Can you finish, analogue way, with the product? Do you know the corresponding formula?