Editing Proof of Bertrand's postulate (section)

===Lemma 4===
An upper bound is supplied for the [[primorial]] function,
:<math>n\#=\prod_{p\,\le\,n}p,</math>
where the product is taken over all ''prime'' numbers <math>p</math> less than or equal to <math>n</math>.

For all <math>n\ge1</math>, <math>n\#<4^n</math>.

'''Proof:''' We use [[Mathematical induction#Complete (strong) induction|complete induction]].

For <math>n=1,2</math> we have <math>1\#=1<4</math> and <math>2\#=2<4^2=16</math>.

Let us assume that the inequality holds for all <math>1\le n\le2k-1</math>. Since <math>n=2k>2</math> is composite, we have
:<math>(2k)\#=(2k-1)\#<4^{2k-1}<4^{2k}.</math>

Now let us assume that the inequality holds for all <math>1\le n\le2k</math>. Since <math>\binom{2k+1}{k}=\frac{(2k+1)!}{k!(k+1)!}</math> is an integer and all the primes <math>k+2\le p\le2k+1</math> appear only in the numerator, we have
:<math>\frac{(2k+1)\#}{(k+1)\#}\le\binom{2k+1}{k}=\frac12\!\left[\binom{2k+1}{k}+\binom{2k+1}{k+1}\right]<\frac12(1+1)^{2k+1}=4^k .</math>
Therefore,
:<math>(2k+1)\#=(k+1)\#\cdot\frac{(2k+1)\#}{(k+1)\#}\le4^{k+1}\binom{2k+1}{k}<4^{k+1}\cdot4^k=4^{2k+1}.</math>