Remainder of division of ${6}^{7^n}$ to $43$ 
What is the Remainder of the division of ${6}^{7^n}$ to $43$?

I've tried with Fermat's little theorem, but it haven't work.
Update : lab bhattacharjee gave a nice proof. But I want to know if this proof is correct.
${6^7}^n = ({6^7})^{7^{n-1}}$
$6^7 = ({6^3})^2$ * 6
$6^3 = 216 = 1(mod 43) => ({6^3})^2 = 1^2 = 1 (mod 43) => 6^7 = 6 (mod 43) => ({6^7})^{7^{n-1}} = 6^{7^{n-1}} (mod 43) <=> {6^7}^n = {6^7}^{n-1} (mod 43)$ 
The same way we obtain $6^{7^{n-1}} = 6^{7^{n-2}} (mod 43)$ 
$6^{7^{n-2}} = 6^{7^{n-3}} (mod 43)$
...
${6^7}^2 = {6^7}^1 (mod 43)$
${6^7}^1 = {6^7}^0 = 6 (mod 43)$
So ${6^7}^n = 6 (mod 43)$

As $6^3\equiv1\pmod{43}$  $$6^{(7^n)}\equiv6^{(7^n)\pmod3}\pmod{43}$$
and $7\equiv1\pmod3\implies7^n\equiv1^n\equiv1$
$$6^{(7^n)}\equiv6^1\pmod{43}$$
