I wish to prove that $\frac{n!}{p_1!p_2!p_3!....p_m!}$ is an integer, were, $p_1+p_2+p_3+...+p_m=n$ and $p_i, n\in \mathbb{N}$. Pleace do check the validity of my proof.
Let us consider the following multinomial, $(x_1+x_2+x_3+...+x_m)^n$. We know that the coefficient of the term: $x_1^{p_1}x_2^{p_2}x_3^{p_3}...x_m^{p_m}$ is $\frac{n!}{p_1!p_2!p_3!....p_m!}$. As the coefficient of each $x_i$ is and integer, thus the coefficient of $x_1^{p_1}x_2^{p_2}x_3^{p_3}...x_m^{p_m}$ must also be an integer. Thus, $\frac{n!}{p_1!p_2!p_3!....p_m!}\in \mathbb{N}$.
I am also open to take better proofs, but do tell me if this is valid enough.
I'd suggest some combinatorial argument.
Consider a set of $n$ balls, where among them, there are $m$ kinds of balls.
And there are $p_i$ number of balls of type $i$, i.e. $p_1$ number of balls of $1$-st kind, $p_2$ number of balls of $2$-nd kind, and so on, like $p_m$ number of balls of $m$-th kind.
Moreover, each kind of balls are identical, where different kinds of balls are distinguishable. For clarification, all the $p_i$ balls of $i^{th}$ type are identical, but all the $p_i$ balls are distinct form all the $p_j$ balls, whenever $i\neq j$.
Now, in how many ways you can arrange the balls?
It is $$\frac{n!}{p_1!p_2!\dots p_m!},$$ because, as each of the $p_i$ balls are identical among itself, we need not consider them as different count. So, divide thee number of permutations of each of those $p_i$ balls, which is $p_i!$.
Now, note that, number of ways ways, can never be a fraction. So, it is necessarily an integer.
Thus, $\frac{n!}{p_1!p_2!\dots p_m!}$ is an integer.