Numbers Equal to Number of Digits in Factorial

Theorem

For $n \in \set {1, 22, 23, 24}$ the number of digits in the decimal representation of $n!$ is equal to $n$.

First we note that:

$1! = 1$

which has $1$ digit.

$2! = 2$

which has $1$ digit.

Multiplying $n$ by a $1$-digit number increases the number of digits of $n$ by no more than $1$.

Thus from $n = 3$ to $n = 9$, $n!$ has no more than $n - 1$ digits.

$21! = 51 \, 090 \, 942 \, 171 \, 709 \, 440 \, 000$

and so $21!$ has $20$ digits.

Multiplying $n$ by a $2$-digit number increases the number of digits of $n$ by at least $1$.

Thus from $n = 10$ to $n = 20$, $n!$ has no more than $n - 1$ digits.

It is noted that:

$22! = 1 \, 124 \, 000 \, 727 \, 777 \, 607 \, 680 \, 000$

which has $22$ digits.

$23! = 25 \, 852 \, 016 \, 738 \, 884 \, 976 \, 640 \, 000$

which has $23$ digits.

$24! = 620 \, 448 \, 401 \, 733 \, 239 \, 439 \, 360 \, 000$

which has $24$ digits.

$25! = 15 \, 511 \, 210 \, 043 \, 330 \, 985 \, 984 \, 000 \, 000$

which has $26$ digits.

As above, multiplying $n$ by a number with $2$ digits or more increases the number of digits of $n$ by at least $1$.

So if $n!$ has more than $n$ digits, $\paren {n + 1}!$ has more than $n + 1$ digits.

It follows by induction that for all $n \ge 25$, $n!$ has more than $n$ digits.

Hence the result.

$\blacksquare$