Write $34560=2^8*3^3*5$. If 6 integers are given, two will have the same remainder mod 5. Their difference is a multiple of 5 and we are done for the exponent of 5. Now the 6 numbers can be either even or odd. If 5 or more numbers have the same parity, the corresponding differences are multiples of 2, and there are more than 8 differences. Hence the number is divisible by $2^8$ in this case. Now we consider the remaining cases: 4 of one parity, 2 of the other parity: Using mod 4 argument, one can get that the 7 differences which are obtained by taking differences of numbers of same parity will be a multiple of $2^8$ because there will exist certain pairs with difference which is a multiple of 4. 3 odd,3 even: Again mod 4 argument solves this case. Take the 6 differences obtained by taking 2 even or 2 odd numbers and prove that their product is a multiple of $2^8$. Hence the exponent of 2 is taken care of. Now we prove for the exponent of 3. The numbers can be 0,1,2 mod 3. If there are 3 numbers with the same remainder, take the 3 products with 2 numbers having same remainder mod 3 so the product of the differences will be a multiple of $3^3$. Otherwise there will be 2 numbers 0 mod 3, 2 numbers 1 mod 3 and 2 numbers 2 mod 3. Multiply the 3 differences which are multiples of 3. Easily this will be a multiple of $3^3$. Hence the product of all possible differences of pairs of 6 numbers is divisible by $2^8*3^3*5=34560$.

We can prime factorize 34560 into $2^8*3^3*5$. Lets split this up into proving that our product has the desired number of each prime factor: Subcase 1: Has a prime factor of 5: By Pigeonhole principle, there exists at least two of our given integers which have the same residue mod 5. Thus, there does indeed exist one prime factor of 5. Subcase 2. Has 3 3's: If there are at least 3 of our numbers that have the same residue mod 3, then we are done. The other case we have to consider is if each of the residues have 0, 1, and 2 correspond to 2 numbers. If this is the case, then each of the three pairs of the two numbers with the same residue mod 3 correspond to having a difference that is a multiple of 3. Hence, this case is done. Subcase 3: Has 8 2's We will take cases on how many of our given six numbers are multiples of 2: 5 or 6 numbers are multiples of 2: We are done 4 numbers are multiples of 2: We have to break into further subcases on this: If at least two of the numbers are multiples of 4, then we are done. If one number or no numbers is a multiple of four, then we are done. 3 numbers are multiples of 2: We have to break into subcases here, and we find that this case works as well. 2 numbers are multiples of 2: We can do similar subcases to our 4 case. 1 number is a multiple of 2: We are done No numbers are multiples of 2: Then we are done