Theorem 2, by Lee Morgenstern, December 2007.
In a 6x6 pandiagonal multiplicative magic square, the magic product is a 6th power.

The proof is in 2 parts.
Part 1 proves that the magic product must be a square.
Part 2 proves that the magic product must be a cube.
Together, they prove that the magic product must be a 6th power.

Part 1

a c a c a c
- b - b - b
a c a c a c
- b - b - b
a c a c a c
- b - b - b

Let P be the magic product.
Let A, B, and C be the products of the a-squares,
the b-squares, and the c-squares, respectively.

The product of the 3 rows of a-squares and c-squares
A x C = P^3.
The product of the 3 columns of b-squares and c-squares
B x C = P^3.
Thus we have
A = B.

The product of the 3 pan-diagonals of a-squares and b-squares
A x B = P^3,
but since A = B, this is the same as
A^2 = P^3,
which is true only if P is a square.

Part 2

a a d a a d
c b - c b -
b c - b c -
a a d a a d
c b - c b -
b c - b c -

Let P be the magic product.
Let A, B, C, and D be the products of the a-squares,
the b-squares, the c-squares, and the d-squares, respectively.

The product of the 2 rows of a-squares and d-squares,
A x D = P^2.
The product of the 2 pan-diagonals of b-squares and d-squares
B x D = P^2.
The product of the 2 pan-diagonals of c-squares and d-squares
C x D = P^2.
Thus we have
A = B = C.

The product of the 4 columns of a-squares, b-squares, and c-squares
A x B x C = P^4.
but since A = B = C, this is the same as
A^3 = P^4,
which is true only if P is a cube.

QED

You can also use the diagrams above for a 6x6 additive pan-magic square, proving that the magic sum, S, must be a multiple of 6, using addition in place of multiplication.
In Part 1, you would get 2A = 3S, proving that S is a multiple of 2.
In Part 2, you would get 3A = 4S, proving that S is a multiple of 3.

BTW, the reason that there is no normal 6x6 pan-magic square is that the normal magic sum of 1...36 divided by 6 is 111, an odd number, but an even number is required to be pan-magic.