Using proof by contradiction, show that $e^z$ is the only entire function that satisfies the conditions $f'(z) = f(z)$ and $f(0) = 1$
I'm stuck on finding the contradiction:
Suppose that $f(z)$ is an entire function, is not equal to $e^z$, and satisfies the given conditions
Let $f(z) = u(x,y) + iv(x,y)$
Then $f'(z) = u_x(x,y) + iv_x(x,y)$
And because $f'(z) = f(z)$, we see that $u = u_x$
However, I do not know if this is the best way to approach the problem, and if it is, what I can derive from this.