Prove that $\frac{\left(\sum_{i = 1}^rm_{i} - 1\right)! \cdot \gcd(m_1, m_2, \cdots, m_{r - 1}, m_{r})}{\prod_{i = 1}^rm_{i}!} \in \mathbb Z^+$. 
Given naturals $m_1, m_2, \cdots, m_{r - 1}, m_r$, prove that $$\large \frac{\displaystyle \left(\sum_{i = 1}^rm_{i} - 1\right)! \cdot \gcd(m_1, m_2, \cdots, m_{r - 1}, m_{r})}{\displaystyle \prod_{i = 1}^rm_{i}!} \in \mathbb Z^+$$

We have that $$v_n(m!) = \sum_{i = 1}^{+\infty}\left\lfloor\frac{m}{n^i}\right\rfloor$$
It is sufficient to prove that $$\left\lfloor\frac{\displaystyle \sum_{i = 1}^rm_{i} - 1}{n}\right\rfloor \cdot v_n(\gcd(m_1, m_2, \cdots, m_{r - 1}, m_{r})) \ge \sum_{i = 1}^r\left\lfloor\frac{m_i}{n}\right\rfloor$$
Write $m_i = np_i + q_i$ where $p_i \in \mathbb Z^+$ and $0 \le q_i < n$ $(i = \overline{1, r})$.
We have that $$\sum_{i = 1}^rm_{i} - 1 = n \cdot \sum_{i = 1}^rp_i + \sum_{i = 1}^rq_i = np + q$$
where $$p = \sum_{i = 1}^rp_i + \left\lfloor\frac{\displaystyle \sum_{i = 1}^rq_i}{n}\right\rfloor$$ and $$q = n \cdot \left\{\frac{\displaystyle \sum_{i = 1}^rq_i}{n}\right\}$$
$(p \in \mathbb Z^+$ and $0 \le q < n)$
The above inequation becomes $$n \cdot \left\{\frac{\displaystyle \sum_{i = 1}^rq_i}{n}\right\} \cdot v_n(\gcd(m_1, m_2, \cdots, m_{r - 1}, m_{r})) \ge \sum_{i = 1}^r\left\lfloor\frac{q_i}{n}\right\rfloor$$
Then I'm not sure what to do next.
 A: Write $d=\mathrm{gcd}(m_1,\ldots,m_r),\, A= (\sum_{i=1}^r m_i )-1,$ and $ B= \prod_{i = 1}^rm_{i}!$.
What you need to show is $$ v_p(A) + v_p(d)\geq v_p(B)$$
for all primes $p$, where $v_p(m)= \lfloor\log_p{m}\rfloor = \sum_{j = 1}^\infty\left\lfloor\frac{m}{p^j}\right\rfloor$.
It suffices to show that $$\left\lfloor\frac{ A }{n}\right\rfloor + \left\lfloor\frac{d}{n}\right\rfloor\ge \sum_{i = 1}^r\left\lfloor\frac{m_i}{n}\right\rfloor.$$
Using your notation, this becomes $$ \sum_i p_i +\left\lfloor \frac{(\sum_i q_i)-1}{n}\right\rfloor + \left\lfloor\frac{d}{n}\right\rfloor\ge \sum_i p_i.$$
If there is $q_i>0$ , then the last inequality is trivially true; Otherwise $q_i=0$ for all $i$, in which case $n$ divides $d$. So $\left\lfloor\frac{d}{n}\right\rfloor\ge  1$, and  the inequality holds again.
A: Your claim is equivalent to the statement that for all integers $r>1$ given any $m_1,\ldots m_r\in\mathbb{N}$ satisfying $\small m_1+\cdots +m_r=n$ the multinomial coefficient ${{n}\choose{m_1,m_2,\ldots,m_{r}}}$ is divisible by $\frac{n}{\gcd(m_1,\ldots m_r)}$.
To prove this note for every integer $1\leq k\leq r$ that$\binom{n}{m_{k}}$ divides ${{n}\choose{m_1,m_2,\ldots,m_{r}}}$ further if we define $g_k=\gcd(n,m_k)$ then by Bezouts lemma we see $\frac{n}{g_k}$ divides $\binom{n}{m_{k}}$  thus  $\frac{n}{g_k}$ divides ${{n}\choose{m_1,m_2,\ldots,m_{r}}}$ again for every integer $1\leq k\leq r$ therefore this means the integer $L=\text{lcm}(\frac{n}{g_1},\ldots \frac{n}{g_r})$ divides ${{n}\choose{m_1,m_2,\ldots,m_{r}}}$ yet since $\text{lcm}(\frac{n}{g_1},\ldots \frac{n}{g_r})=\frac{n}{\gcd(g_1,\ldots g_r)}=\frac{n}{\gcd(m_1,\ldots m_r)}$ this completes the proof.
