# $p_n$ is never a divisor of $n!$

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:

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

• Since $p_2=3$ you need not exclude $n=2$. – gammatester Sep 23 '18 at 21:48
• 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 – Adam Sep 23 '18 at 21:50
• IMO it is simpler with $p_n>n.$ – gammatester Sep 23 '18 at 21:50
• 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? – Adam Sep 23 '18 at 21:53
• 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? – gammatester Sep 23 '18 at 22:03