Bertrand-Chebyshev Theorem/Lemma 2

Lemma for Bertrand-Chebyshev Theorem
For all $m \in \N$:


 * $\ds \prod_{p \mathop \le m} p \le 2^{2 m}$

where the product is taken over all prime numbers $p \le m$.

Proof
It is plainly true for $m \le 2$; we will proceed by strong induction.

If $m > 2$ is even then:

By Sum of Binomial Coefficients over Lower Index:


 * $\ds \sum_{r \mathop = 0}^{2 k + 1} \binom {2 k + 1} r = 2^{2 k + 1}$

Therefore:

If $m = 2 k + 1$ is odd, then all the prime numbers $k + 2 \le p \le 2 k + 1$ divide:


 * $\dbinom {2 k + 1} k = \dfrac {\paren {2 k + 1}!} {k! \paren {k + 1}!}$

Thus: