Arithmetic puzzle

[houlahop]

22!=1124000727777607680000
The number 22! is a 22 digit number

11240
00727
77760
76800
00

Find a number n such as n! is n^2 digit number

More generally find n such as n! is n^k digit number (k>2)

[jaap]

It is not possible.

For n>1 we have:
log n! < log nⁿ = n log n < n²

So n! always has fewer than n² digits (except for 1! and 0!).

[houlahop]

Thank you

(n^2)/d(n!) is equal to pi(n) when n goes to infinity (where pi(n) is the counting function of primes)

Is there any interpretation of this "equality"?

My last post because I wanted to point out to this.
Good luck and good bye!
To the moderator :
No one asked me about clarification. Sir Gabriel answered without asking.
Conclusion : Either you did not read the post, either you are lying.
Where did you see any contempt from myself?

[SecondTalon]

Alright then. Bye.

[Schrödinger's Wolves]

Thank you

(n^2)/d(n!) is equal to pi(n) when n goes to infinity (where pi(n) is the counting function of primes)

Is there any interpretation of this "equality"?

My last post because I wanted to point out to this.
Good luck and good bye!
To the moderator :
No one asked me about clarification. Sir Gabriel answered without asking.
Conclusion : Either you did not read the post, either you are lying.
Where did you see any contempt from myself?

I would guess that interpretation is lim(n->infinity) (n^2)/(d(n!)*pi(n))=1.
What is the function d in this case?

[Xias]

I believe d(n) is just shorthand for "the number of digits in n."

I wonder about the fact that n^2 and pi(n) are not dependent on the base, but d(n) is.