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For infinite products, a nonzero limiting product requires that its factors tend to one. A useful sufficient condition isAfter finitely many factors, $|u_j|<1/2$ and the principal holomorphic logarithm satisfies $|\log (1+u_j)|\le 2|u_j|$. Therefore the sum of logarithms converges, and exponentiating it gives a finite nonzero product. The same argument on compact sets proves infinite product convergence from logarithmic tails: a locally uniformly absolutely convergent tail of holomorphic logarithms gives a holomorphic nonvanishing tail product. Finite factors then determine all zeros and their orders.

The Weierstrass elementary factors areFor $|w|<1$,Their only zero is a simple zero at $w=1$.

Work with entire functions on $\mathbb{C}$. As usual for prescribed exact zero orders, the distinct zero locations must have consistent multiplicities. The literal statement allows repeated locations with conflicting orders; that cannot be true, for example if the same point is prescribed order one and order two. Remove consistent repetitions rather than adding their orders, and separate a possible zero at the origin.

List the distinct nonzero locations as $a_j$, with prescribed orders $n_j$, and write $m$ for the prescribed order at zero, or zero if the origin is not prescribed. Choose $p_j$ large enough thatThen the Weierstrass factorization theorem construction isOn every compact set, $|z/a_j|\le 1/2$ for all sufficiently large $j$ because $|a_j|\to \infty$. Its logarithmic tail is bounded byThus the product converges locally uniformly, is entire, and has exactly the specified zeros with exactly their orders. At a prescribed zero, only its own finite factor vanishes; all other finite factors and the tail product are nonzero.

Without escape to infinity the conclusion fails in general. Distinct proposed zeros $1/j$ accumulate at zero. The identity theorem would force an entire function with those zeros to vanish identically, which does not have precisely the prescribed isolated zeros. A finite zero set can of course be realized by a polynomial.

For entire functions with the same zero divisor, the quotient $q=f/g$ extends through every common zero by cancelling equal local powers. It is entire and nowhere zero. Hence $q^{\prime }/q$ is entire and has a primitive on the simply connected plane. Choose $h(0)$ with $e^{h(0)}=q(0)$ and putThe derivative of $q{e}^{-h}$ vanishes, and its value at zero is one. ThereforeIf also $f=e^{k}g$, then $e^{h-k}=1$ everywhere. The continuous difference $h-k$ takes values in the discrete set $2\pi i\mathbb{Z}$ and so is constant on the connected plane:The whole-plane hypothesis matters: a zero-free holomorphic quotient on a multiply connected domain need not possess a global holomorphic logarithm.

A nonzero entire function can be described by its discrete zero locations and multiplicities, a product of Weierstrass elementary factors and a zero-free exponential factor. Conversely, consistent prescribed multiplicities at distinct points escaping every compact subset can be realized by choosing factor orders large enough for infinite product convergence from logarithmic tails. Repeated locations with incompatible exact multiplicities are not admissible data, and a finite accumulation of distinct zeros is excluded by the identity theorem.