We first note that an equivalent statement to Bertrand's postulate is $\frac{(2n)!}{n!}$ contains a prime factor, where it is JUST a prime. Not a power of a prime. Just p. What I did was this: Denote $v_p(n!)$ be the maximum value of the exponent of p dividing n!. We also use the fact that $v_p(n!)$=$\sum_{i=1}^{\infty}$ $\frac{n}{p^{i}}$=S. (I can't put floor functions sorry). Then we have that $\frac{(2n)!}{n!}$=${p^{S}}$A. So we can set ${p^{S}}$=(n+$\epsilon$). By the fundamental theorem of arithmetic, $\frac{(2n)!}{n!}$ has a prime decomposition. Continuing on this process for another prime, say q, we have $\frac{(2n)!}{n!}$=$p^{S}$$q^{S'}$B. We also notice that S' $\leq$ S, and $q^{S'}$=(n+$\delta$). Doing this over and over again, we have the descending chain of inequalitites, $S^{''...'}$ $\leq$ ...S'. So eventually, one must have that there is some exponent $S^{''...'}$, and $p^S{''...'}$=(n+l), such that $S^{''...'}$=1, by (induction, or some staircase argument?...) and we are done. In particular, there is a p in $\frac{(2n)!}{n!}$.
The problem for me here, is that one could use this argument for anything. But, if (2n)! gets too small, there is not even a prime factorization. I guess that was my argument. I don't know if I am right, but it seems to very trivially work. (Of course one could do this for $\frac{(2n+\epsilon)!}{n!}$, but I think one could do this for smaller than (2n)!).