show that $a^{12} \equiv 1 \mod 32760$ for each $a \in \mathbb{Z}$ such that $\text{gcd}(a, 32760) = 1$ I am trying to show that $a^{12} \equiv 1 \mod 32760$ for each $a \in \mathbb{Z}$ coprime to $32760$. I have found the prime decomposition $32760 = 2^3 \cdot 3^3 \cdot 5 \cdot 7 \cdot 13$ and determined $\phi(32760)$, where $\phi$ is the Euler's totient function. This gives me that $\phi (32760) = 8640$. 
Using Euler's congruence, this gives that $a^{8640} \equiv 1 \mod 32760$, but this is still far from the result I want to show... I also know a consequence of Eulers congruence, which states that $e \equiv e' \mod \phi (n)$ implies that $a^e \equiv a^{e'} \mod n$, but I do not see how to use this...
Any hints would be appreciated.
$\textbf{EDIT:}$ I made a mistake in my primefactorization, it is edited now.
 A: We have
$$32760 =  2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13$$
By the Chinese remainder theorem,
$$U(32760)
\cong U(8) \times U(9) \times U(5) \times U(7) \times U(13)
\cong C_2 \times C_2 \times C_6 \times C_4 \times C_6 \times C_{12}$$
which has exponent $12$.
In terms of congruences, by Euler–Fermat we have
$\quad a^2 \equiv 1 \bmod 8$
$\quad a^6 \equiv 1 \bmod 9$
$\quad a^4 \equiv 1 \bmod 5$
$\quad a^6 \equiv 1 \bmod 7$
$\quad a^{12} \equiv 1 \bmod 13$
and so $a^{12} \equiv 1 \bmod 8,9,5,7,13$
By the Chinese remainder theorem,
$\quad a^{12} \equiv 1 \bmod 32760 $
A: For each maximal prime power $\,p^{\large k}$ dividing $\, 32760 = 2^{\large 3} \cdot 3^{\large 2} \cdot 5\cdot 7\cdot 13\,$ we have that Euler's $\,\phi = \phi(p^k)\mid 12,\ $ so $\,12 = \phi k,\,$ so $\  {\rm mod}\,\ p^{\large k}\!:\,\ a^{\large 12} \equiv (\color{#c00}{a^{\large \phi}})^{\large k}\equiv \color{#c00}1^{\large k}\equiv 1\, $ by $\rm\color{#c00}{Euler's}$ Theorem.
Thus, since all $\,p^{\large k}$ divide $\, a^{\large 12}\!-1,\,$ so too does their lcm = product $= 32760$
Remark $\ $ See Carmichael's Lambda Theorem for the general result.
