# $\prod_{ p\leq x}p\leq 4^{x-1}$ for all real $x\geq2$

How you prove this? I'm looking the Erdös proof from Bertrand Postulate and there are many things I don't get. Please don't hints, I'm newbie in combinatorics techniques.

In the book I don't get how $$\displaystyle\prod_{m+1< p\leq 2m+1}p \leq \left( \begin{matrix} 2m+1 \\ m \end{matrix} \right).$$

$\left( \begin{matrix} 2m+1 \\ m \end{matrix} \right) = \dfrac{(m+2)(m+3)(m+4)\cdots(2m+1)}{m!}$
Note that if p is a prime such that $m+1< p \leq 2m+1$, then $p$ is a factor of the numerator but not the denominator. Hence $p \mid \left( \begin{matrix} 2m+1 \\ m \end{matrix} \right)$. Hence, so too does the product of all such prime numbers.