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.



searching for an elementary proof a complex analysis result
Let $L$ be your difference operator: $(Lf)(z)=f(z+1)-f(z)$.
Consider these polynomials
$$P\_n(z)=\frac{1}{n!}z(z-1)\ldots(z-n+1),\quad n=0,1,2,\ldots.$$
Simple computation shows that $LP\_n=P\_{n-1}$. 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.


