Very long rithm

We just need to apply the laws of logarithm to solve this question.

$log_{100!}2+log_{100!}3+.............+log_{100!}99+log_{100!}100$ can be written as:

$log_{100!}(2\times3\times....\times99\times100)$ (since $log_{a}(b)+log_{a}(c) = log_{a}(bc)$ - Law of Logarithm)

$log_{100!}(2\times3\times....\times99\times100) = log_{100!}(100!)$ (since $100!=2\times3\times....\times99\times100$ )

$log_{100!}(100!)=1$ (since $log_{a}(a)=1$ - Law of Logarithm)

Thus, the answer is: $log_{100!}100!=\boxed{1}$

$log_{a}(b)+log_{a}(c)=log_{a}(bc)$

therefore, as the listed logs of course multiply to 100! due to the definition of the factorial function, we get $log_{100!}(100!)$ which of course is $\boxed{1}$