# Is it true for $n > 2$ then there always exists a prime $\le n$ that does not divide $n$?

I was thinking of how to prove $\frac{n^n}{n!}$ is never an integer for $n > 2$. I think if I prove the above question, then this follows immediately.

• Since $\gcd(n,n-1)=1$, no prime factor of $n-1$ divides $n$. And since $n>2$, there must exist a prime dividing $n-1$. – Levent Apr 22 '16 at 20:28
• @Levent, brilliant answer... – L.F. Cavenaghi Apr 22 '16 at 20:30
• @Levent Great! Should have thought of that! If you write an answer, I will accept it. – taninamdar Apr 22 '16 at 20:33

Since $\gcd(n,n-1)=1$, no prime factor of $n-1$ divides $n$. And since $n>2$, there must exist a prime dividing $n-1$.