Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$. 
Show that $a^{13} \equiv a \pmod{3 \cdot 7 \cdot 13}$.

I want to know if my attempt is correct.
First $a^{13} \equiv (a^3)^4 \cdot a \equiv a^4 \cdot a \equiv a^3 \cdot a^2 \equiv a \cdot a^2 \equiv a^3 \equiv a \pmod 3$.
Second, $(a^7)^2 \cdot a^{-1} \equiv a^2 \cdot a^{-1} \equiv a \pmod 7$.
Third, $a^{12} \equiv 1 \pmod{13} \implies a^{13} \equiv a \pmod{13}$.
So, $3, 7, 13 | a^{13} - a$. Since they are relatively prime in pairs, $$3 \cdot 7 \cdot 13 | a^{13} - a \implies a^{13} \equiv a \pmod{ 3 \cdot 7 \cdot 13}$$
Is it correct? Could you suggest an easier or a better proof?
Thanks.

By Euler's Theorem (or you may even use Fermat's theorem),
$a^2 \equiv 1 \pmod {3} \implies a^{12} \equiv 1 \pmod 3 \implies a^{13} \equiv a \pmod 3$
$a^6 \equiv 1 \pmod {7} \implies a^{12} \equiv 1 \pmod 7 \implies a^{13} \equiv a \pmod 7$
$a^{12} \equiv 1 \pmod {13} \implies a^{13} \equiv a \pmod {13} $
$\therefore a^{13} \equiv a \pmod {lcm[3,7,13]} \implies a^{13} \equiv a \pmod {3\times7\times13}$
It is quit trivial if $a$ is not co-prime to at least one of $3, 7, 13.$ Then you will not even need Euler or Fermat's theorem.

Remember, if
$a\equiv b \pmod {m_1}$
$a\equiv b \pmod {m_2}$
.
.
.
$a\equiv b \pmod {m_n}$
then, $a\equiv b \pmod {lcm[m_1,m_2,...,m_n]}$

