$a^{13} \equiv a \bmod N$ - proof of maximum $N$ From Fermat's Little Theorem, we know that $a^{13} \equiv a \bmod 13$. Of course $a^{13} \equiv a \bmod p$ is also true for prime $p$ whenever $\phi(p) \mid 12$ - for example, $a^{13} = a^7\cdot a^6 \equiv a\cdot a^6 = a^7 \equiv a \bmod 7$.
So far I have that the largest $N$ for which all $ a^{13} \equiv a \bmod N$, is $N = 2\cdot 3\cdot 5\cdot 7\cdot 13 = 2730$
Can someone either put together an elegant proof of this, or find and prove a different limit?

Putting in $a=2$, we get that $N$ divides $2^{13} - 2 = 2 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13$. 
On the other hand, putting in $a=3$, we get that $N$ divides $3^{13} - 3 = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 73$. 
Hence $N$ must divide $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 = 2730$. 
