What is the remainder of ${{6457}^{76}}^{57}$ modulo $23$? What is the remainder when ${{6457}^{76}}^{57}$ is divided by $23$?

How to solve it by Euler's theorem or Chinese theorem?
 A: $6457\equiv-6\pmod{23}$
$\implies6457^{76^{57}}\equiv(-6)^{76^{57}}\pmod{23}\equiv6^{76^{57}}$
Now we need $76^{57}\pmod{\phi(23)}$
As $(22,76)=2,$ let us find $76^{57-1}\pmod{22/2}$
Now $76\equiv-1\pmod{11}\implies76^{56}\equiv(-1)^{56}\equiv1$
$76^{57}=76\cdot76^{56}\equiv1\cdot76\pmod{11\cdot76}\equiv76\pmod{22}\equiv10$
$\implies6^{76^{57}}\equiv6^{10}\pmod{23}$
A: $\ \ \ ca\bmod cn\,=\, c\,(a\bmod n)\ $ as we explained here, hence
$ 76^{\large 57}\!\bmod 22\, =\, 2\,(38\cdot 76^{\large 56}\bmod 11)\, =\, 2\,(5(-1)^{\large 56}) = 10\ $ so $\ \color{#c00}{76^{\large 57}\! =10\! +\! 22k}$
${\rm mod}\,\ 23\!:\,\ 6457^{\large{ 76^{\Large 57}}}\!\!\! \equiv17^{\large \color{#c00}{76^{\Large 57}}}\!\!\!\equiv17^{\large\color{#c00}{ 10+22k}}\equiv 17^{\large 10}{\underbrace{(17^{\large 22})^{\large k}\equiv 1^{\large k}}_{\rm Fermat}}17^{\large 10}\equiv \color{#0a0}{17^{{\large 10}}}\ $   
A: By Euler's Theorem, $\varphi(23)=22$ and applying Fermat's Theorem $1\equiv 6457^{\varphi(23)} \pmod{23}$
$10 \equiv 76 \pmod{\varphi{(23)}}$
$7 \equiv 57 \pmod{\varphi({\varphi{(23)}})}$
$10 \equiv 10^{7} \pmod{\varphi{(23)}}$
So we know that $6457^{76^{57}}\equiv 6457^{10} \pmod{23}$
$17\equiv 6457 \pmod{23}$
Thus $4\equiv 6457^{76^{57}} \pmod{23}\equiv 6457^{10} \pmod{23} \equiv 17^{10} \pmod{23}$
