When will D(n!) and D(D(n!)) reach a googol or googolplex?

[bomba923]

Let's say you have a digit function, D(n), which equals the number of digits in the input (n). (e.g., 19 becomes 2, 142151 become 6...well, you see)
(where n! is, well, a factorial of a factorial)

1) Now when will D(n!) become a googol? A googolplex?
2) Now when will D(D(n!)) become a googol? A googolplex?
3) In subjective but ""somewhat"" reasonable language, how fast do you think the sequence S(n)=D(n!)-n increases? (slow..med..fast...whichever way to describe it)

(For easy calculation of digits, just set your graphing calculator to "Scientific" exponential format in the MODE screen)

 

[matt grime]

You realize that D(n) is just floor{log(n)}+1, where I'm taking logs base 10.

 

[bomba923]

You realize that D(n) is just floor{log(n)}+1, where I'm taking logs base 10.

EXAcTLY! This was an AMC question somewhere , but I liked this problem because I couldn't find the method to solve the Double factorial that would quickly yield the solution in 2 to 5 minutes (it used to be on a test, about 2-5 minutes per problem)--anyway, the factorial I could take of, but the double factorial got more difficult; i tried expanding it as a sum of logs, but (until I wasted way 2much time for that) then it became more difficult; so, I was wondering what would be not only a method, but a 2 to 5 minute technique that found conceptually would have allowed to solve it, without going back and forth.
The last question, about [D(n!)]-n, I was curious upon observation of the sequence graph; it appears as a concave upward shape with minimums of -3 for 'n' from 6 to 14 inclusively. But after that, it actually seemed to expand quite slowly (again, using the unfortunately "subjective" language again ) Until finally, I found that it bounded a sequence from below of a function previously found during the solving of this AMC problem. I thought it cool,
but how would I solve the double factorial issue with a two to five minute technique?? i.e., what conceptual path should be followed here?
The googol-googolplex issue was added for reasons unknown, (except maybe just to annoy me! )

 

[NateTG]

Well there's Stirling's formula
$$n! \approx \sqrt{2\pi}n^ne^{-n}$$
http://mathworld.wolfram.com/StirlingsApproximation.html

 

[bomba923]

Well there's Stirling's formula
$$n! \approx \sqrt{2\pi}n^ne^{-n}$$
http://mathworld.wolfram.com/StirlingsApproximation.html

Thanx! I'll try just that