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Here is a simple proof of your divisibility relation. It suffices to show that $$\frac{\mathrm{lcm}(1,2,\dots,n+j)}{\mathrm{lcm}(1,2,\dots,n)}\quad\text{divides}\quad\binom{n+j}{n}\quad\text{for}\quad 0\leq j\leq n.$$ That is, for any prime $p$ and for $0\leq j\leq n$, we have that $$\lfloor\log_p(n+j)\rfloor-\lfloor\log_p(n)\rfloor\leq\sum_{k=1}^\infty \left(\left\lfloor\frac{n+j}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor-\left\lfloor\frac{j}{p^k}\right\rfloor\right).$$ Note that the terms on the right hand side are nonnegative integers. In addition, for $\lfloor\log_p(n)\rfloor<k\leq\lfloor\log_p(n+j)\rfloor$, the $k$-th term is positive, because in this case $n+j\geq p^k>n\geq j$. Hence the right hand side is at least the number of $k$'s satisfying $\lfloor\log_p(n)\rfloor<k\leq\lfloor\log_p(n+j)\rfloor$, and we are done.