The intersection always contains the product, so we only need show the product contains the intersection.
Let's get started with two ideals. Let H1 be P1 primary and let H2 be P2 primary. Since P1 and P2 are maximal, write x1 + x2 = 1, where x1 is in P1 and x2 is in P2. Choose n so that x1n is in H1 and x2n is in H2, and raise the equation to the 2n.
(x1 + x2)2n = 1
Expand the left hand side using the binomial theorem. A linear combination of elements from H1 and H2 spans 1.
Let z lie in the intersection of H1 and H2. Multiply through by z, and a linear combination of elements from H1 and H2, times z, yields z. Thus z is contained in the product ideal H1H2. The intersection is contained in the product, and the intersection equals the product.
Now proceed by induction on the number of ideals. Assume this holds for n ideals, and let J be the intersection, or product, of the underlying primary ideals. Introduce a new P1 primary ideal H1. Write n equations that span 1.
x1 + y1 = 1
x2 + y2 = 1
x3 + y3 = 1
…
All the values x1 through xn come from P1. The values y1 through yn come from the remaining n prime ideals. Multiply these equations together, and the only term on the left not in P1 is the product of y1 through yn, which is in the product of the other prime ideals. By induction, the product equals the intersection, and the intersection of these prime ideals is equal to rad(J). Something in P1, plus something in rad(J), equals 1.
u + v = 1
Raise this equation to a high power, and apply the binomial theorem, and H1+J spans 1.
Select any z in the intersection, and multiply through by z. When z is multiplied by an element from H1, treat z as an element in the product of the remaining n primary ideals. The result lies in the product of all n+1 primary ideals. When z is multiplied by something in J, remember that z lies in H1. This too lies in the product of all n+1 primary ideals. Put this all together and the product spans z, which is an arbitrary element in the intersection. That completes the inductive step, and the proof.