We break down properties of numbers.

What are your multiples of 5?

5, 10, 15, 20, 25, 30, 35, 40, 45, 50

Out of these, two are multiples of 25 (25, 50) so they contain two “5”s each

The remaining eight numbers contain one “5” each, for a total of 12 “5”s altogether.

There are at least 25 “2”s in the list. However, each “2” pairs with one “5” to form a zero.

With there only being 12 “5”s, we can only form a number which ends with 12 zeroes.

Can list the 12 pairs? Tks

By pairing I mean pairing a 2 with a 5. You cannot "list them out" at all.

How many "2"s are there? More than 25.
How many "5"s are there? Only 12.

Each "2" pairs with a "5" to form a 10, since 2 x 5 = 10.

You can only form 12 of such pairs since you are limited by the number of "5"s. Each pair will contain one "zero" in the resulting product.

So, you can only have 12 "10"s altogether, resulting in 10 x 10 x 10 x 10 x ... (12 of them altogether) = a number with 12 zeroes behind,

If you really wish to spell the numbers out, this is what happens.

NOTE: We do not count 10, 20, 30, 40 and 50 as having zeroes behind; for the purposes of this question, I will break 10, 20, 30, 40 and 50 into their "2"s and "5"s.

6 = 2 x ... (contains one "2")
8 = 2 x 2 x 2 (contains three "2"s)
10 = 2 x ... (contains one "2")
12 = 2 x 2 x ... (contains two "2"s)
14 = 2 x ... (contains one "2")
16 = 2 x 2 x 2 x 2 (contains four "2"s)
18 = 2 x ... (contains one "2")

So far, there are 13 "2"s altogether. Obviously there's more.

In short, we do likewise for 5, 10, 15, ..., 50 and found that there are 12 "5"s altogether.

So, 1 x 2 x 3 x 4 x 5 x ... x 50
= 2 x 2 x 2 x 2 x 2 x ... x 5 x 5 x 5 x 5 x 5 x ... x random numbers not containing 2 and 5 as any of their factors (such as "3", "7" and so on)
= 10 x 10 x 10 x ... x other numbers
= number with 12 "10"s

(But we are not going to calculate the entire multiplication literally)

Well explained, tks