, 2023

If $n!$ ends with exactly
$m$ zeroes, then $n!$ is divisible by $10^m$ but not divisible by $10^{m+1}$. (If $n!$ were divisible by $10^{m+1}$, it would end with at least
$m+1$ zeroes.)

In this case, we can write $n! = 10^m $\cdot$
q$where$q$ is not divisible
by 10. This in turn means that either $q$ is not divisible by 2 or not divisible
by 5 or both.

Since $2<5$, when $n \geq 2$, the product $n! = 1 $\cdot 2 \cdot 3 \cdot \cdots \cdot$ (n-1)
$\cdot$ n includes more multiples of 2 than of 5 among the n$integers in its product, so$n!$ includes more factors of 2 than of
5.

This in turn means that, if $n!$ ends
in exactly $m$ zeroes, then $n! = 10^m \cdot q$ where $q$ is not divisible by 5, and so the
number of zeroes at the end of $n!$
is exactly equal to the number of prime factors of 5 in the prime
factorization of $n!$.

We note also that as $n$ increases,
the number of zeroes at the end of $n!$ never decreases since the number of
factors of 5 either stays the same or increases as $n$ increases.

For $n=1$ to $n=4$, the product $n!$ includes 0 multiples of 5, so $n!$ ends in 0 zeroes.

For $n=5$ to $n=9$, the product $n!$ includes 1 multiple of 5 (namely 5),
so $n!$ ends in 1 zero.

For $n=10$ to $n=14$, the product $n!$ includes 2 multiples of 5 (namely 5
and 10), so $n!$ ends in 2
zeroes.

For $n=15$ to $n=19$, the product $n!$ includes 3 multiples of 5 (namely 5,
10 and 15), so $n!$ ends in 3
zeroes.

For $n=20$ to $n=24$, the product $n!$ includes 4 multiples of 5 (namely 5,
10, 15, and 20), so $n!$ ends in 4
zeroes.

For $n=25$ to $n=29$, the product $n!$ includes 5 multiples of 5 (namely 5,
10, 15, 20, and 25) and includes 6 factors of 5 (since 25 contributes 2
factors of 5), so $n!$ ends in 6
zeroes.

For $n=30$ to $n=34$, $n!$ ends in 7 zeroes. For $n=35$ to $n=39$, $n!$ ends in 8 zeroes.

For $n=40$ to $n=44$, $n!$ ends in 9 zeroes. For $n=45$ to $n=49$, $n!$ ends in 10 zeroes.

For $n=50$ to $n=54$, $n!$ ends in 12 zeroes, since the product
$n!$ includes 10 multiples of 5, two
of which include 2 factors of 5.

For $n=55$ to $n=74$, $n!$ will end in 13, 14, 15, 16 zeroes as
$n$ increases.

For $n=75$ to $n=79$, $n!$ ends in 18 zeroes.

For $n=80$ to $n=99$, $n!$ ends in 19, 20, 21, 22 zeroes as $n$ increases.

For $n=100$ to $n=104$, $n!$ ends in 24 zeroes.

For $n=105$ to $n=124$, $n!$ ends in 25, 26, 27, 28 zeroes.

For $n=125$, $n!$ ends in 31 zeroes since $125$ includes 3 factors of 5, so $125!$ ends in 3 more zeroes than $124!$.

Of the integers $m$ with $1 \leq m \leq 30$, there is no value of
$n$ for which $n!$ ends in $m$ zeroes when $m = 5, 11, 17, 23, 29, 30$, which means
that $30 - 6 = 24$ of the values of
$m$ are possible.