matinyousefi wrote: Prove that $\binom{n+k}{n}$ can be written as product of $n$ pairwise coprime numbers $a_1,a_2,\dots,a_n$ such that $k+i$ is divisible by $a_i$ for all indices $i$. More simply put, we want to show that: For each prime $p$ dividing the product $(k+1)(k+2)...(k+n)$, the following relation holds: $$v_p(n!) \ge \sum v_p(k+i) - \max _{1 \le j \le n}(v_p(k+j))$$ Indeed, if $\max _{1 \le j \le n}(v_p(k+j))$ holds for $j=m$, it suffices to distribute the $p$-powers in $n!$ through the multiples of $p$ in $k+1,...,k+n$ different of $k+m$ to "cancel" their factors. This way, prime $p$ will divide only $a_m$. Now we prove our claim. Lets say that $v_p(k+m)=d$. Looking to the other factors $k+i$, we see that $$v_p((k+m)-(k+i)) \ge v_p(k+i)$$ Therefore, $v_p((k+1)...(k+m-1)(k+m+1)...(k+n) \le v_p((m-1)(m-2)...1.1...(n-m)) = v_p((m-1)!)+v_p((n-m)!)$ But by binomials, the RHS is less or equal to $v_p((n-1)!)$, which proves our result.