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I saw the question $f$ is entire without any zeros then there is an entire function $g$ such that $f=e^g$

I understand that for any non-vanishing entire function $f(z)$:

If there exists an entire function $g$ such that $f(z)=e^{g(z)}$, then $f'(z)=g'(z)\times f(z)$.

But how should I prove the reverse implication, which seems to be correct?

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It is not exactly correct, but almost.

Set $h(z)=f'(z)/f(z)$. Then $h(z)$ is an entire function. Let $H(z)$ be some anti-derivative of $h(z)$.

Now compute the derivative of $f(z)/\exp(H(z))$. You will see it is $0$.

Thus $f(z)/\exp(H(z))$ is constant, and you basically have what you want.

You then still have the possibility of choosing the anti-derivative in such a way that you get exactly what you need.

For example you could impose that the antiderivative at $H(0) = \log f(0)$ (where $\log$ is some branch of the logarithm it does not matter).

Put differently a correct statement along your lines is:

"For $f$ entire and non-zero, if $f'(z)= g'(z) f(z)$ and $g(0) = \log f(0)$, then $f(z) = e^{g(z)}$."

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