Determine whether $\frac{1000!}{100!^{10}}$ is an integer Can you give an idea, how to find out whether the result of ${1000!}/{100!^{10}}$ an integer. 
Modulo division? But what I met was about powers like $2^{100}/125$...
 A: Consider the binomial coefficents ${200 \choose 100},{300 \choose 100},{400 \choose 100},{500 \choose 100},{600 \choose 100},{700 \choose 100},{800 \choose 100},{900 \choose 100},{1000 \choose 100}$
For example, 
$${700 \choose 100} = \frac{700!}{600!100!} = \frac{700\cdot 699\ldots 602\cdot 601}{100!}$$
and this is an integer... 
A: Here’s a non-combinatorial, more number-theoretic proof. For any prime $p$ the number of factors of $p$ in $n!$ is
$$\sum_{k\ge 0}\left\lfloor\frac{n!}{p^k}\right\rfloor\;,$$
so it suffices to show for an arbitrary prime $p$ that
$$10\sum_{k\ge 0}\left\lfloor\frac{100}{p^k}\right\rfloor\le\sum_{k\ge 0}\left\lfloor\frac{1000}{p^k}\right\rfloor\;.\tag{1}$$
$(1)$ will certainly be true if
$$10\left\lfloor\frac{100}{p^k}\right\rfloor\le\left\lfloor\frac{1000}{p^k}\right\rfloor\tag{2}$$
for primes $p$ and $k\ge 0$. But
$$\left\lfloor\frac{100}{p^k}\right\rfloor\le\frac{100}{p^k}\;,$$
so
$$10\left\lfloor\frac{100}{p^k}\right\rfloor\le\frac{1000}{p^k}\;,$$
and $(2)$ (and hence $(1)$) follows immediately.
A: The basic result is that the product of $k$ consecutive numbers is divisible by $k!$.
This is the fact that ${{n}\choose{k}}$ is an integer when $n \ge k$.
Now, $1000!$ can be decomposed into $10$ products of $100$ consecutive numbers.
A: What's the number of ways to split $1000$ people into $10$ teams of $100$ ?
A: Since you tagged this abstract algebra, I will give a hint based on it: the group $$\underbrace{S_{100}\times\cdots\times S_{100}}_{10\text{ copies}}$$ has an obvious embedding as a subgroup of $S_{1000}$.
A: I have a stronger statement:   
$$\binom{1000}{100}=\frac{1000!}{100!^{10}\binom{900}{100}\binom{800}{100}\binom{700}{100}\binom{600}{100}\binom{500}{100}\binom{400}{100}\binom{300}{100}\binom{200}{100}}$$   
is an integer. More generally:    
$$\binom{ab}{b}=\frac{(ab)!}{b!^{a}\binom{(a-1)b}{b}\binom{(a-2)b}{b}\cdots\binom{2b}{b}}$$   
is an integer (for $a\ge 2, b\ge 0$).
