# Find the number of digits in n!.

number of digits in n = floor(log(n)) + 1;

so for n! # of digits = (floor(log(n!)))+1;

= floor(Sum of log1+log2+--- + logn)+1;

digits = (int) log10(number), where log 10 means log for base 10

--

♥ ¸¸.•*¨*•♫♪♪♫•*¨*•.¸¸♥