Laborious maths failure
 

Now let's put the denominator to one side and think about the numerator. For each of the 15 fractions we need to multiple it's current numerator by everything in the denominator we just calculated other than by anything already in the denominator of that fraction. We need to then multiple each of those 15 large products together into one huge product.

So we get:
[1*(2^4)*(3^5)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[29*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*83]
*[7*(2^3)*(3^6)*(7^2)*11*13*19*29*41*43*79*83*89
*[9*(2^4)*(3^6)*(7^2)*(11^2)*13*19*41*43*79*83*89]
*[13*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*79*83*89]
*[5*(2^3)*(3^4)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[2*(2^4)*(3^6)*7*(11^2)*13*19*29*41*43*79*83*89]
*[23*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*89]
*[11*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*43*79*83*89]
*[7*(2^4)*(3^3)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[1*(2^2)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[19*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*83*89]
*[3*(2^4)*(3^6)*(7^2)*(11^2)*19*29*41*43*79*83*89]
*[17*(2^4)*(3^6)*7*11*13*19*29*41*43*79*83*89]
*[4*(2^4)*(3^6)*(7^2)*(11^2)*13*29*41*43*79*83*89]

Getting rid of the two 1's, which do not change the final product, and breaking down the two composite numbers (4 and 9) into prime factors, we get:
[(2^4)*(3^5)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[29*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*83]
*[7*(2^3)*(3^6)*(7^2)*11*13*19*29*41*43*79*83*89
*[(3^2)*(2^4)*(3^6)*(7^2)*(11^2)*13*19*41*43*79*83*89]
*[13*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*79*83*89]
*[5*(2^3)*(3^4)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[2*(2^4)*(3^6)*7*(11^2)*13*19*29*41*43*79*83*89]
*[23*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*89]
*[11*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*43*79*83*89]
*[7*(2^4)*(3^3)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[(2^2)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[19*(2^4)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*83*89]
*[3*(2^4)*(3^6)*(7^2)*(11^2)*19*29*41*43*79*83*89]
*[17*(2^4)*(3^6)*7*11*13*19*29*41*43*79*83*89]
*[(2^2)*(2^4)*(3^6)*(7^2)*(11^2)*13*29*41*43*79*83*89]

Grouping like prime factors together inside each part, and getting everything into numerical order within each part, gives us:
[(2^4)*(3^5)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[(2^4)*(3^6)*(7^2)*(11^2)*13*19*(29^2)*41*43*79*83]
*[(2^3)*(3^6)*(7^3)*11*13*19*29*41*43*79*83*89
*[(2^4)*(3^8)*(7^2)*(11^2)*13*19*41*43*79*83*89]
*[(2^4)*(3^6)*(7^2)*(11^2)*(13^2)*19*29*41*79*83*89]
*[(2^3)*(3^4)*5*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[(2^5)*(3^6)*7*(11^2)*13*19*29*41*43*79*83*89]
*[(2^4)*(3^6)*(7^2)*(11^2)*13*19*23*29*41*43*79*89]
*[(2^4)*(3^6)*(7^2)*(11^3)*13*19*29*43*79*83*89]
*[(2^4)*(3^3)*(7^3)*(11^2)*13*19*29*41*43*79*83*89]
*[(2^2)*(3^6)*(7^2)*(11^2)*13*19*29*41*43*79*83*89]
*[(2^4)*(3^6)*(7^2)*(11^2)*13*(19^2)*29*41*43*83*89]
*[(2^4)*(3^7)*(7^2)*(11^2)*19*29*41*43*79*83*89]
*[(2^4)*(3^6)*7*11*13*17*19*29*41*43*79*83*89]
*[(2^6)*(3^6)*(7^2)*(11^2)*13*29*41*43*79*83*89]

Now we are ready to simplify all 15 parts of this huge product by together by grouping together the powers of each prime factor, which gets us:
(2^59)*(3^87)*5*(7^32)*(11^29)*(13^15)*17*(19^15)*23*(29^15)*(41^14)*(43^14)*(79^14)*(38^14)*(89^14)