Prove $a {a+b \choose b}$ divides the lowest common multiple of $b+1, b+2, ..., b+a$ Prove that for $a, b \in \Bbb N^*$ we have :
$$a {a+b \choose b} | LCM(b+1, ..., b+a)$$
Is there a simple proof of this result that doesn't involve computing an integral? I know a trick exploiting $\int_{0}^{1} {x^n(1-x)^mdx}$ but I am looking for a more "obvious" proof.
 A: First note that 
$$a\binom{a+b}{b}=\frac{(a+b)!}{(a-1)!b!}$$
To prove that $a {a+b \choose b} | LCM(b+1, ..., b+a)$ it is enough to prove that for each prime $p$ we have 
$$p^n |  LCM(b+1, ..., b+a)$$
where $n$ is the power of $p$ in $a\binom{a+b}{b}$.
We know by Legendre formula that 
$$n=\sum_{k}\lfloor \frac{a+b}{p^k} \rfloor -\lfloor \frac{b}{p^k} \rfloor-\lfloor \frac{a-1}{p^k} \rfloor$$
Now for each $k$ we have
$$\lfloor \frac{b}{p^k} \rfloor+\lfloor \frac{a-1}{p^k} \rfloor \leq  \frac{b}{p^k} + \frac{a-1}{p^k} \leq \frac{a+b}{p^k}  $$
We also have
$$\frac{a}{p^k} \leq \lfloor \frac{a-1}{p^k} \rfloor +1\\
\frac{b}{p^k} <\lfloor \frac{b}{p^k} \rfloor +1$$
Combining, we get 
$$\lfloor \frac{b}{p^k} \rfloor+\lfloor \frac{a-1}{p^k} \rfloor  \leq \frac{a+b}{p^k} < \lfloor \frac{b}{p^k} \rfloor+\lfloor \frac{a-1}{p^k} \rfloor +2$$
which implies 
$$\lfloor \frac{a+b}{p^k} \rfloor -\lfloor \frac{b}{p^k} \rfloor-\lfloor \frac{a-1}{p^k} \rfloor \in \{ 0,1 \} $$
Now, let $K_0$ be the largest $k$ such that 
$$\lfloor \frac{a+b}{p^k} \rfloor -\lfloor \frac{b}{p^k} \rfloor-\lfloor \frac{a-1}{p^k} \rfloor =1 $$
Show that this implies that $p^{K_0}$ has a multiple between $b+1$ and $b+a$.
This shows that $p^{K_0}$ divides $LCM(b+1,..,b+a)$. Now, since all terms in the sum are 0 or 1, it follows that $n \leq K_0$ and hence $p^n | p^{K_0}$.
