Question

Question #db134

Answer 1

About `3.455*10^9014`

Explanation:

This can be caluculated at
http://www.wolframalpha.com/input/?i=product+of+n%5En+from+n%3D1+to+n%3D100

This type of sequence is called hyperfactorial.

https://oeis.org/A002109
http://googology.wikia.com/wiki/Hyperfactorial

[Approximation]
Let the product `P=prod_(n=1)^100n^n=1^1*2^2* * * 100^100`.
Then,
`ln P=sum_(n=1)^100nlnn` and
`int_1^100xlnxdx< ln P < int_1^101xlnxdx`
(See about Riemann sum: https://en.wikipedia.org/wiki/Riemann_sum)

Since `intxlnxdx=(x^2lnx)/2 -x^2/4 +C`,
`[(x^2lnx)/2 -x^2/4]_1^100 < ln P < [(x^2lnx)/2 -x^2/4]_1^101`
`5000ln100 - 2499.75 < lnP < 5100.5ln101 - 2550`
`20526.1 < ln P < 20989.4`
`e^20526.1< P < e^20989.4`
`2.355*10^8914 < P < 3.807*10^9115`

The geometric mean between the lower bound and the upper bound is
`sqrt(2.355*10^8914* 3.807*10^9115)=9.469*10^9014`
and this is a bit larger than the exact value of P.