the Hadamard factorization theorem
/ Hadamard -> ah-dah-MAR /
A polynomial of degree n factors completely: p(z) = c times the product of (z - a_k) over its n roots, and once you know the roots and the leading constant you know the whole polynomial. Can an entire function be reconstructed from its zeros the same way? Almost. The Weierstrass factorization theorem already says yes in principle, but it leaves the structure loose. Hadamard's theorem is the sharp version for functions of finite order: it pins down exactly how the product over the zeros must be built and exactly what extra exponential factor is allowed, with both pieces dictated by the order rho.
The statement: let f be entire of finite order rho, with a zero of order m at the origin and other zeros a_1, a_2, ... (with multiplicity). Then f(z) = z^m times e^(g(z)) times the canonical product P(z) over the zeros, where two things are controlled by rho. First, the canonical product P(z) is built from Weierstrass elementary factors E_p(z/a_k) with genus p equal to the integer part of rho (more precisely p is the smallest integer with the zeros' exponent sum convergent, and p <= rho). Second, g(z) is a polynomial of degree at most rho. So the 'mystery' exponential factor e^(g) is not arbitrary — its degree is bounded by the order. Reading this backward: the order, the zeros' density, and the factorization are three views of one fact. For example, an order-1 function with infinitely many zeros has g(z) of degree at most 1, i.e. g(z) = a z + b, and a genus-0 or genus-1 product.
Hadamard's theorem is the structural crown of entire-function theory. It powers the proof of the little Picard theorem, it lets you identify functions from their zeros (it is how one proves the product formula for sin z, with g of degree 0 and a genus-1 product), and it underlies the analysis of the Riemann zeta and gamma functions. The decisive caveat: it requires finite order. A function like e^(e^z) has infinite order, no Hadamard form, and lies entirely outside this clean picture — for such functions the looser Weierstrass factorization is all you get.
Apply Hadamard to sin(pi z). It is entire of order 1 with zeros at every integer. The theorem forces g(z) of degree at most 1 and a genus-1 canonical product. Matching the value of the derivative at 0 pins g to a constant, yielding the famous formula sin(pi z) = pi z times the product over n >= 1 of (1 - z^2 / n^2). The order dictated the whole shape; only one constant had to be fixed by hand.
Hadamard forces the form of sin(pi z) from its order and zeros — the product formula is essentially inevitable.
Hadamard needs finite order; for infinite order only the weaker Weierstrass factorization survives. And the genus of the product and the degree of g are both bounded by — not necessarily equal to — the order, so reading them off requires care.