No prob. Sorry if I go on at too much length here ;-)
So a useful formula for counting digits is, a positive integer x has floor(log10(x)) + 1 digits. You can figure this out yourself by thinking, where does the digit-counting function bump up a notch? At 10, 100, 1000, etc. So asymptotically the number of digits in x is O(log x).
So n! has O(log n!) digits. The trickier part is figuring out or knowing that O(n log n) = O(log n!). Using log ab = log a + log b you can expand out the factorial:
In case the last step isn't clear, you can do this splitting-in-half bounding trick. Since each element in the sum is less than log n you can bound from above with
log n + log (n-1) + ... + log 2 + log 1 < n log n
And if you just take the larger half of the list you can bound from below with
Usually, it seems to be either (n log n) or (n log n) - 1 digits.
And usually in this case I would leave of the O, as that usually refers to the performance of an algorithm in time or space. I suppose you could construe the number of digits to be "space" but multiplying O(n) numbers doesn't make that much sense.