Hey guys so I am reading through a proof and I am completely stuck on one part. The lemma is $n\# < 4^n$ where $n\#$ is the primorial function. So far I have understood a bit of the proof and it goes something like this Proof: We show this by induction. The cases $n = 1, 2$ work. Now assume $k\# < 4^n$. If n is not prime, then $n\# = (n-1)\#$ and so $n\# = (n-1)\# <= 4^{n-1} < 4^n$. Now we wish to show the inductive case for n prime. Since n>2, it is odd and we may write $n = 2m + 1$
Now this is the part I get stuck on. Notice that ${2n+1 \choose n}$ is divisible by all prime numbers greater than $n+1$ and less than or equal to $2n+1$, that is, it is divisible by $\frac{(2n+1)\#}{(n+1)\#}$. Now how exactly do we get to this jump?
Consider the integer $c = {2n+1 \choose n} = {(2n+1)! \over n!(n+1)!}$. A prime $p$ in the range $(n+1, 2n+1)$ divides the numerator but not the denominator. So $p$ must divide $c$.