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I just wanted to know if this will suffice as proof, as the title suggests, I want to prove that $p_n$ is never a divisor of $n!$ unless $n=2$, and I feel like stating QED after making the following statement regarding the p-adic valuation of $n!$ for the $n^{th}$ prime, but yes this is very brief and so I need criticism of course:

$$\sum _{j=1}^{ \bigl\lfloor {\frac {\ln \left( n \right) }{\ln \left( p_{{n}} \right) }} \bigr\rfloor +1} \Bigl\lfloor {\frac {n}{{p _{{n}}}^{j}}} \Bigr\rfloor =0 \,\,\,\,\forall n: n \gt 1 \land n \in \mathbb N $$

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    $\begingroup$ Since $p_2=3$ you need not exclude $n=2$. $\endgroup$ – gammatester Sep 23 '18 at 21:48
  • $\begingroup$ Ah right of course that was silly I just panicked for a minute and threw that in because I do often make that mistake, stating something to apply for a primes and I should exclude 2 $\endgroup$ – Adam Sep 23 '18 at 21:50
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    $\begingroup$ IMO it is simpler with $p_n>n.$ $\endgroup$ – gammatester Sep 23 '18 at 21:50
  • $\begingroup$ Sure lol I get what you are saying but suppose for a moment that such an obvious inequality does not exist, would such an approach be considered suffice for proof? $\endgroup$ – Adam Sep 23 '18 at 21:53
  • $\begingroup$ If you compute some sums, you will notice that they actually boil down to $$\sum _{j=1}^{ 1} \Bigl\lfloor {\frac {n}{{p _{{n}}}^{j}}} \Bigr\rfloor = \Bigl\lfloor {\frac {n}{{p _{{n}}}}} \Bigr\rfloor =0 $$ So how do you prove the general sum formula? $\endgroup$ – gammatester Sep 23 '18 at 22:03

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