If $n = ab$ where $a \ne b$, then why do $a$ and $b$ divide $(n-1)!$ ? I've seen this in a few proofs, such as Prove that if $n$ is composite, then $(n-1)! \equiv 0 \pmod n$. It makes sense logically, except that I can't seem to make sense mathematically.
I know that $(n-1)! = 1 \times 2 \times 3 \times ... \times (n-1)$. But, how would I know $a$ and $b$ are in the product of integers $\le (n-1)$?
Edit: I should add, I also know that if $n=ab$, where $a \ne b$, that $a$ and $b$ are $<n$.
The original statement is incorrect: $3=1\cdot3$, where $1\ne 3$, but $3$ does not divide $(3-1)!$. The correct statement is that if $n=ab$, where $1<a<n$, then $a\mid(n-1)!$ and $b\mid(n-1)!$. In this case $a$ is one of the integers $2,3,\ldots,n-1$, so $a$ is one of the factors of
$$(n-1)!=1\cdot 2\cdot 3\cdot\ldots\cdot (n-1)\;.$$
Moreover, $1<a<n$ implies that $n=\frac{n}1>\frac{n}a>\frac{n}n=1$, and $\frac{n}a=b$, so $1<b<n$, and the same argument applies to $b$.
You just proved it yourself! Suppose $n = ab$ with $a$ and $b$ integers and further, $a \not = b$.
$a$ is surely smaller than $n$ right? Thus it is one of the numbers $1, 2, \cdots, n-1$. We need to impose $a < n $.
The same train of thought applies to $b$, $b$ is one of the numbers $1, 2, \cdots, n-1$. Similarly we have $b < n $ and therefore $1 < a, b < n $.
What is more, $a$ and $b$ are different, therefore you can multiply $1, 2, \cdots, n-1$ out and remove both $a$ and $b$ from the product, still getting a product of some numbers below $n$. Therefore $ab$ divides $(n-1)!$
A more visual representation, assuming $a < b$:
$$(n-1)! = (n-1)\cdot(n-2)\cdot\ldots\cdot b\ldots\cdot a\cdot\ldots\cdot2\cdot1$$
and you can easily see that you can remove both $a$ and $b$ from it, thus you can divide $(n-1)!$ by $n$.
