The Taylor series for e^x is sum[k=0 to inf] (x^k)/(k!).
Substituting x=-1 means 1/e = e^(-1) = sum[k=0 to inf] ((-1)^k)/(k!)
Then 100000!/e can be written as 100000! * ( sum [k=0 to inf] ((-1)^k)/(k!) )
The first 100000 terms (k=0 to 99999) can be written as (-1)^k * 100000 * 99999!/k!.
99999!/k! is an integer for integer k from 0 to 99999 which then means that all those terms are integer multiples of 100000.
The 100001st term (k=100000) is (-1)^100000 * 100000!/100000! = 1.
The sum of the first 100001 terms will be an integer, any further terms will be fractional. This integer ends with the five digits 00001.
This series is a convergent alternating series, so the error of the sum of the first n terms is bounded by the (n+1)th term.
In this case the error is the 100002th term (k=100001), which is (-1)^100001 * 100000!/100001! ~= -0.00000999990000099999....
This is so small it does not affect the nearest integer, so the last 5 digits of [100000!/e] are 00001.
Solution
