Prove that $~~~~~a_1!a_2!\cdots a_m! \mid \left(a_1+a_2+...+a_m\right)! ~~~~~~~~~\forall ~~~a_1,a_2,...,a_m\in N$

Question

Show that $$a_1!a_2!\cdots a_m! \mid \left(a_1+a_2+...+a_m\right)! \forall a_1,a_2,...,a_m\in N$$

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Case 1 $m=2$. We need to prove $$a_1!a_2!\mid (a_1+a_2)!$$

1. $a_1+a_2=1$ and $a_1+a_2=2$. It is obvious

2. $a+b=n (n\in \text{N and } a+b\le n).$

We will prove in $$a+b=n+1$$

By assuming of the induction $$(a_1!(a_2-1)!)\mid(a_1+a_2-1)!$$

And $$(a_2!(a_1-1)!)\mid(a_1+a_2-1)!$$

$$\rightarrow (a_1!a_2!)\mid (a_1+a_2)(a_1+a_2-1)!$$

Or $$\rightarrow (a_1!a_2!)\mid (a_1+a_2)!$$

Case 2 $m=k$

We will prove in $m=k+1$

I am stuck here. Can I prove it as with the case $m = 2?$, help me.

Answer 1

By the $m=2$ case: $$ a_1!(a_2+\ldots+a_{n+1})!\,\vert\, (a_1+a_2+\ldots+a_{n+1})!, $$ and by the inductive hypothesis for $m=n$, $$ a_2!\ldots a_{n+1}!\vert (a_2+\ldots+a_{n+1})!. $$ Putting these together gives the $n+1$ case.