Algebraic functions are polynomials? Does any one how to prove that every entire algebraic function is a polynomial? I'm under the impression that this can be achieved by showing that an algebraic function grows no faster than a polynomial. 
 A: I assume you mean every entire algebraic function is a polynomial.  Suppose $g$ is an entire algebraic function of order $n$.  Thus there are polynomials $c_j(z)$, $j=0 \ldots, n$ with $c_n$ not identically $0$ such that $\sum_{j=0}^n c_j(z) g(z)^j = 0$.  For all but finitely many complex numbers $w$, $\sum_{j=0}^n c_j(z) w^j$ is not identically $0$, and so there are at most finitely many $z$ for which $\sum_{j=0}^n c_j(z) w^j = 0$: those are the only $z$ for which we can have $g(z) = w$.  Thus for all but finitely many $w$, $g(z)$ takes the value $w$ only finitely many times.  But an entire function that is not a polynomial has an essential singularity at $\infty$, and by the Great Picard Theorem it takes all but one value infinitely many times.
A: Hopefully not, because $\: \operatorname{exp} : \mathbb{C} \to \mathbb{C} \:$ defined by $\:\:\:\: f(z) \:\: = \:\: \displaystyle\sum_{n=0}^{\infty} \: \left(\frac1{n!} \cdot \left(z^n\right)\right)$

is an entire function that is not a polynomial.

See http://en.wikipedia.org/wiki/Exponential_function#Complex_plane.
