Edit:
I found a proof by P. Erdös in "A theorem of Sylvester and Schur".
Any suggestion to prove one of this problems?
The following problems are equivalent:
Prove that for all naturals $n, m : 1 \le n \le m$ then exists a prime $p > n$ such that: $$p \, |\,\prod_{k = 1}^{n}(m+k)$$
Prove that for all naturals $n, m : 1 \le n \le m$ then: $$ \sum_{p\,\text{prime}}\left\lfloor \binom{n + m}{m} \frac{n!}{p} \right\rfloor - \left\lfloor \frac{n!}{p} \right\rfloor > 0 $$
Prove that for all naturals $n, m : 1 \le n \le m$ then: $$ \omega\left(\prod_{k = 1}^{n}(m+k)\right) > \pi(n) $$
where $\omega$ is the Prime omega function and $\pi$ is the Prime counting function.
By Bertrand's postulate, for $m = n,\ n+1,\ n+2,\ n + 3$ this is true.
Let $\Psi(x,y)$ the number of y-smooth numbers less than or equal to x.
Prove that for all naturals $n, m : 1 < n \le m$ then: $$\Psi(m + n, n) - \Psi(m, n) < n$$
As a corollary of the Størmer's theorem exists a natural $M_n$ such that for all $m \ge M_n$ this is true.
Remark: I don't know if this is true or not, just saw the first one on internet.
Any idea? Is a conjecture writted in another way or something?