# does $ab$ divide $(ab-1)!$?

i am interested in following thing, that for $a \gt 2$ , $b \gt2$, $ab$ divides $(ab-1)!$ ? I can take some simple example, for example $(3,3)$,or $(3,5)$ and show that this is true by this way; but I need ways to proof it by induction or by algebra maybe,how can i do it?
$$(ab-1)! =1\times 2\times 3\times 4\times\cdots\times(ab-1)$$
my logic is that, because we are trying to calculate $(ab-1)!$, then we can find some pair of numbers, with by multiply together will cancel $ab$, because each $a$ and $b$ are less then result of this factorial,this is just a human logic,how to proof it by mathematics logic?

-
What you call "just a human logic" is actually perfectly fine "mathematics logic." –  anon Jan 30 '12 at 21:58

Without loss of generality, let $a\leq b$.
First, consider the case where $a<b$. Since $a,b>2$ ,$ab >2a$ and consequently $ab-1 \geq a$. The same argument shows that $ab-1 \geq b$. So, $a$ and $b$ will both occur in the product $(ab-1)!=1\times 2\times3\times \cdots \times a\times \cdots \times b\times (ab-1)$ for all $a,b > 2$.
The case when $a=b$ is much more subtle. Since $a=b$, $(ab-1)!=(a^2-1)!=1\times 2\times3\times \cdots \times a\times \cdots (ab-1)$, so we know that $a$ divides $(a^2-1)$. We use the hypothesis that $a>2$ to deduce that $a^2>2a. \implies a^2-1 \geq 2a$. So $2a$ is also in the product $1\times 2\times3\times \cdots \times a\times \cdots (ab-1)$. Also note that $a$ is distinct from $2a$ for all non-zero integers. So, use this to summarize that $a^2|(a^2-1)!$.
The argument is fine if $a\lt b$, but runs into a bit of difficulty if $a=b$, no? –  Arturo Magidin Jan 30 '12 at 22:00
@dato. Oh, you dont need to do that case. Just switch the roles of $a$ and $b$ and apply our theorem. This works because $ab=ba$ in the integers. –  Galois Group Jan 30 '12 at 22:17