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?

Observe that $$a^m(a^n-1)\equiv0\pmod9$$ for any integer $a$ if $m\ge2$ and $6\mid n$
So, for integer $k\ge1,$ $$a^{6^{k+1}}-a^6=a^6(a^{6^k}-1)\equiv0\pmod9$$ 
$$\implies\sum_{k=1}^na^{6^k}\equiv na^6\pmod9$$
Here $n=?$
