I'm having trouble getting started on the following:
Given $n_1, n_2, ..., n_k \in$ $\Bbb N$, show that $n_1!\cdot n_2!\cdot\cdot\cdot n_k! |(n_1+n_2+...+ n_k)!$
I thought about a proof by induction on $k$ but feel that induction shouldn't be needed. Ideas?
On
Imagine you have $k$ different letters. The first letter is available $n_1$ times, the second one $n_2$ times etc. Now show that the number of words which you can form out of these $n_1 + n_2 + \ldots + n_k$ letters is $\frac{(n_1 + n_2 + \ldots + n_k)!}{n_1!\cdot\ldots\cdot n_k!}$. In particular, this number is an integer, so $(n_1 + n_2 + \ldots + n_k)!$ is divisible by $n_1!\cdot\ldots\cdot n_k!$.
The ratio $$\frac{(n_1+n_2+\cdots+n_k)!}{n_1!n_2!\cdots n_k!}$$ is a multinomial coefficient. It is the number of words of length $n_1+\cdots+n_k$ with $n_1$ occurrences of the "letter" $a_1$, $n_2$ occurrences of $a_2$, and so on. Since our fraction counts something, it must be an integer.