Problem on exponential of entire function

I am stuck at a problem which says :

" Let $f$ and $g$ are entire functions such that $e^f, e^g$ and $1$ are linearly dependant over $\mathbb{C}$. Show that $f$, $g$ and 1 are also linearly dependant."

So basically we have constants $C_1$, $C_2$, $C_3$ in $\mathbb{C}$, not all zero, such that $C_1 e^f + C_2 e^g + C_3=0$ . It becomes easy if one of the $C_i$ 's is zero. But I am stuck in the case where all of them are non-zero.

I can understand that it suffices to show that $|Af + Bg|$ is bounded on $\mathbb{C}$ for some complex constants $A, B$, not both of them zero. I tried using the inequality $|e^z|\leq e^{|z|}$, but it is not working.

Please help with any suggestion as soon as possible :( Thanks in advance !

Assume all the constants are not zero, then we may assume without loss of generality that $e^f=e^g+1$, then you use the Picard theorem, because the range of $e^f$ doesn't contain 0 and 1(because $e^g$ can't be 0). So you get the conclusion that f and g must be constants.

Suppose we have $c_{1} e^{f}+c_{2} e^{g}+ c_{3}=0$ and $c_1\ne0$ then we have $$e^g= \frac{c_3-c_1 e^f}{c_1}$$ and also we have $$ c_{1}f' e^{f}+c_{2}g'e^{g}=0$$ which implies $$\frac{f'}{g'}=\frac{-c_2 e^{g}}{c_1 e^f}$$ Thus we have $$\frac{f'}{g'}=\frac{-c_2(c_3-c_1 e^f)}{c_1^2 e^f}=\frac{b}{e^f}-c$$ for some constants $b$ and $c$. This implies that $\frac{f'}{g'}$ is bounded, Thus, using Liouville's theorem we get that $\frac{f'}{g'}$ is constant that is $$f'=kg'$$ Now by integrating both sides we get that $$f=kg+l$$ which implies that $\{f,g,1\}$ are linearly dependent.