819 B
The factorial part is pretty easy - we cache the factorials from 1 to 9 and then we can calculate facdigitsum(n) in O(log_10(n)).
The problemdescription tells us there are only seven numbers in a loop ({169, 363601, 1454}, {871, 45361}, {872, 45362}).
So every other chain ends with a number mapping to itself. We manually insert these seven numbers in our grid and calculate the rest (where we only need to test if f(n) == f(n-1)).
And we cache ever calculated value of chainlength() in our 1000000-element array. So as soon as we reach an already calculated number we can stop.
This works because there are no loops (except the seven pre-inserted values) and every chain is strictly linear.
Be sure to check out the pre-optimized version of this to see just how much more condensed this version is :)