Prime product pp(n) >/ 19*n for every natural n >/ 7

This wlog  (for  n >/ 7)  can be read independently of wlog

https://aot2025.blogspot.com/2025/09/prime-product-ppn-5n-of-primes-p-4-p-n.html

while it is a continuation of that previous wlog (for  n >/ 5).

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Terminology:

•  N := {1 2 3 ...} -- the set of all natural numbers
•  P := {2  3  5  7  11  13  17 ... } -- the set of all primes
•  pp(x) := prod{ p in P: p \< x}. -- the product of all primes that are  \<  x

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The product of the empty set, i.e. of zero elements, is equal to 1 (by a definition). This starts the list below:

• pp(1) = 1
• pp(2) = 2
• pp(3) = 2*3 = 6
• pp(4) = pp(3) = 6
• pp(5) = 2*3*5 = 30
• pp(6) = pp(5) = 30
• pp(7) = 2*3*5*7 = 210
• pp(10) = pp(9) = pp(8) = pp(7) = 210
• pp(11) = 2*3*5*7*11 = 2310
• pp(12) = pp(11) = 2310
• pp(13) = 2*3*5*7*11*13 = 30030

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Theorem 1    pp(n) >/ 19*n   for every natural  n >/ 7.

Proof:  Apply the table above:

          7 \< n \< 10    ==>    pp(n)  =  210  >  19*10  >/  19*n

and

         11 \< n \< 100   ==>    pp(n)  >/  2310  >  19*100  >/ 19*n

The theorem already holds whenever  7 \< n\< 100. We can finish the proof by doing the induction on   n  starting with  n=100.  Consider an arbitrary natural  n >/ 100  for which the theorem holds. Then we consider  3. cases:

• Case A:   pp(n) = 19*n
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  Then  n  is divisible by arbitrary prime  p \< n  except for p=19;  our  n  is not divisible by  19.  Thus  n+1  is not divisible by any prime  p \< n  except possibly by 19. Of course  n+1  cannot be divided by any prime  p > n+1.  It follows that there exists natural  k  such that  n+1 = 19^k,  or else  n+1  itself is prime.
  
  If  n+1=19^k  then  18 | 19^k-1 = n  hence  3^2 | n.  However  n | pp(n)   while  3^2  does not divide  pp(n) -- a contradiction;  n+1  is not equal to. 19^k.
  
  Thus  n  has to be prime.  Then
  
         pp(n+1) = pp(n)*(n+1) = 19*n*(n+1) > 19*(n+1)
  
  Thus, the theorem holds for  n+1  (under the Case A).
  
  
  
• Case B:   pp(n) = 19*n + r       where  1 \< r < 19
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  Then  19  does not divide  pp(n)   hence  n < 19  --  a contradiction;  Case B  is impossible.
  
  
  
• Case C:   pp(n)  >/  19*(n+1)     (the remaining case)
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  Then  pp(n+1)  >/  pp(n)  >/  19*(n+1)  hence the theorem holds for  n+1 (under the Case C).

Together, the case of  7 \< n \< 100, and Cases A B C  cover  all integers  n >/ 7.  End of Proof

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REMARK:  Experts have proved results infinitely harder(!) than the one above.