I have not been able to think of a way to exactly calculate the leading digits of your huge number without overflow.
But I did think of a risky and potentially error prone way that might approximate this.
There is an algorithm to quickly calculate arbitrary powers using recursive squaring.
Google can help you find that algorithm and an explanation of how it works.
Then imagine at each step of this process we throw away any power of ten that has appeared.
The mathematica function MantissaExponent can be used to do that.
All this tries to keep the calculated value small enough that the calculation does not overflow and fail.
power[base_,exponent_]:=
If[exponent==0,1,
If[EvenQ[exponent],temp=power[base,exponent/2];MantissaExponent[temp*temp][[1]],
temp=power[base,(exponent-1)/2];MantissaExponent[base*temp*temp][[1]]
]
];
MantissaExponent[power[N[96717311574016,100],16777216^4*8^4*7*2]*7^11][[1]]
which returns
0.519477805513753546135142414870187736952679394740473429808918564205
Ignore the leading "0.", it is the digits that follow this that you want to concentrate on. That claims that your huge number probably begins with the digits 519477...
The ",100" in that tells Mathematica you want to try to use 100 digit floating point precision in every step. If you change that to ",200" then it will try to use 200 digits. But before you change this to 200 or 10,000 or 1,000,000 or more, study this and the Google search result explaining the process and then try this on much much smaller, but gradually increasing in size, numbers. Do this with small enough numbers that you can use Mathematica to calculate the exact value and repeat the calculation with this approximate method and see how many of the digits from the two methods agree. I am certain that if you try this on enough examples and the size of the exact result is close to or beyond the number of floating point digits that you ask it to use that you will find examples where some of the the last digits in the result will be incorrect. I am hoping this will let you see some of the leading digits and that enough of those will be correct that you can use them. Be very very careful with this. If anyone finds any mistake in this then please point it out to me. Thank you
