Given a function $ g $ entire on the whole complex plane $ C $, it is possible to find an entire function $f $ such that $ f(z+1) f(z)=g(z) $. The proof can be given using riemann surface,automorphy,covering,etc. Can anyone find a elementary proof which avoids all such things.
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Let $L$ be your difference operator: $(Lf)(z)=f(z+1)f(z)$. Consider these polynomials $$P_n(z)=\frac{1}{n!}z(z1)\ldots(zn+1),\quad n=0,1,2,\ldots.$$ Simple computation shows that $LP_n=P_{n1}$. Polynomials $P_n$ make a basis in the space of all polynmials, because there is one polynomial of each degree. This allows you to find a solution of any equation with polynomial RHS. Then perform a limit process. For the details see any book under the title Calculus of finite differences. For example, by N\"orlund or by Gelfond. 

