Showing that $a^{12}\equiv 1\pmod{210}$ if $a$ is coprime to $210$ I ran into this next question:

Show that if $a$ is coprime to $210$:
$a^{12}\equiv1 \pmod{210}$.

This is not a homework question.
Thank you very much in advance,
Yaron

HINT:
Using Carmichael Function,
$$\lambda(210)=\text{lcm}\{\lambda(2),\lambda(3),\lambda(5),\lambda(7)\}=\text{lcm}(1,2,4,6)=12$$
In fact,we can find a larger $n$ such that $a^{12}\equiv\pmod n$ for all $a$ co-prime to $n$
We need $\lambda(n)$ divides $12$
Case $1: $ If $p=2,$ we know, $\lambda(2^r)=\begin{cases} 2^{r-1} &\mbox{if } 1\le r \le 2 \\
2^{r-2} &  \text{ otherwise }\end{cases}$
$2^{r-2}$ needs to divide $12,r-2\le 2\implies r \le 4$
Case $2: $ If $p$ is odd prime, $\lambda(p^r)=\phi(p^r)=p^{r-1}(p-1),$
If $r>1,$ as the only odd divisor$(>1)$ of $12$ is $3^1\implies p=3,r=2$
Else $r=1\implies p-1$ must divide $12\implies p$ can be $2,3,5,7,13$
$2,3$ have already been encountered 
So, the highest value of $n$ will be $13^1\cdot7^1\cdot5^1\cdot3^2\cdot2^4$
As $a^{12}\equiv1\pmod {13\cdot7\cdot5\cdot3^2\cdot2^4}$ if $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1,$
$a^{12}\equiv1\pmod d$ where $d$ is any divisor of $13\cdot7\cdot5\cdot3^2\cdot2^4$ and $(a,d)=1$ as $(a,13\cdot7\cdot5\cdot3^2\cdot2^4)=1$
