Proof that $\frac{n!}{j_1! j_2! j_3! \cdots j_k!} \in \mathbb{Z}$ if $j_1+ j_2+ j_3+\cdots+ j_k = n$ $n, j_1, j_2, j_3, \ldots, j_k \in \mathbb{N}$ are such that: $j_1+ j_2+ j_3+\cdots+ j_k = n$. Prove that $$\frac{n!}{j_1! j_2! j_3! \cdots j_k!} \in \mathbb{Z}.$$ I don't know how to do it. Tried induction and $e^{\ln(j_1! j_2! j_3! \cdots j_k!)}$ but non of those seem to be working.
 A: The given expression counts the number of ways to order $n$ objects, with $j_1$ objects of type $1$, $j_2$ of type $2$ and so on until $j_k$ objects of type $k$, and there are no other objects or types. Thus the expression must be an integer.
A: For induction on $k$, the base step $k=1$ gives a ratio $\tfrac{n!}{n!}=1$. To go from $k=j$ to $k=j+1$ in the inductive step, replace $\tfrac{1}{j_k^\text{old}!}$ with $\frac{1}{j_k^\text{new}!j_{k+1}!}$, wth $j_k^\text{new}=j_k^\text{old}-j_{k+1}$. This multiplies the ratio by $\frac{j_k^\text{old}!}{j_k^\text{new}!j_{k+1}!}=\binom{j_k^\text{old}}{j_{k+1}}$, which is an integer. In fact, this insight allows us to prove by telescoping product$$\frac{n}{\prod_{j=1}^kj_j!}=\prod_{l=1}^{k-1}\binom{n-\sum_{j=1}^{l-1}j_j}{j_l}.$$
A: Assuming that binomial coefficients are integers (they are, by a simple combinatorial argument), you can use them with telescoping to write your expression:
\begin{align*}
&\binom{j_1+\cdots +j_k}{j_1}\cdot \binom{j_2+\cdots +j_k}{j_2}\cdots \binom{j_{k-1}+j_k}{j_{k-1}}\cdot \binom{j_k}{j_k}\\
    &= \frac{(j_1+\cdots +j_k)!}{j_1!\cdot (j_2+\cdots +j_k)!}\cdot \frac{(j_2+\cdots +j_k)!}{j_2!\cdot (j_3+\cdots +j_k)!}\cdots \frac{(j_{k-1}+j_k)!}{j_{k-1}!\cdot j_k!}\cdot \frac{j_k!}{j_k!\cdot 0!}\\
    &= \frac{(j_1+j_2+\cdots +j_k)!}{j_1!\cdot j_2!\cdots j_k!}
\end{align*}
