Prove: $9\mid a^6+a^{6^2}+\ldots+ a^{6^{27}},\;a\in\mathbb N$ 
Prove $$9\mid a^6+a^{6^2}+\ldots +a^{6^{27}},a\in\mathbb N$$

My attempt:
\begin{aligned}&a \equiv 2\pmod{9}\implies a^{6} \equiv 2^{6} \equiv 1\pmod{9}\\&\begin{array}{l}{a \equiv 3\pmod{9}\implies a^{6} \equiv 3^{6} \equiv 0\pmod{9}} \\{a \equiv 4\pmod{9}\implies a^{6} \equiv\left(2^{6}\right)^{2} \equiv 1\pmod{9}}\\a\equiv 5\equiv -4\pmod{9}\implies a^6\equiv(-4)^6=4^6\equiv 1\pmod{9}\end{array}\end{aligned}
$$a^{6^i}=\left(a^6\right)^{6^{i-1}}\equiv [0,1]\pmod{9}\implies a^{6^i}=9k\;\lor\;a^{6^i}=9k+1,\;k\in\mathbb N_0$$
$$a^6+a^{6^2}+\ldots +a^{6^{27}}=\sum_{i=1}^{27}\left(a^6\right)^{6^{i-1}}=\sum_{i=1}^{27}(9k_i+[0,1])=\underbrace{\sum_{i=1}^{27}9k_i+[0,27]}_{\text{multiple of}\; 9}$$
//edited mistake [27,54]//
Is this correct and is there a more efficient method?

It is correct, but a bit long.
Another method will use Euler's theorem: $\;\varphi(9)=6$ so, if $a$ is not divisible by $3$, we have $a^6\equiv 1\mod 9$, and by an easy induction, $\;a^{6^k}=\bigl(a^6 \bigr)^{6^{k-1}}\equiv 1$. Therefore
$$a^6+a^{6^2}+\ldots+ a^{6^{27}}\equiv\underbrace{1+1+\ldots+1}_{27 \:1\text{s}}\equiv 0\mod 9 $$
for this case.
On the other hand, if $a$ is divisible by $3$, $a^2\equiv 0\bmod 9$, hence $a^6\equiv 0$ too.
