Anti Fermat's little theorem

We can apply fermat's little theorem to each of the prime divisors:- So we have $\displaystyle 2^{6}\equiv 1\mod7$

$\displaystyle 2^{12}\equiv 1 \mod 13$

$\displaystyle 2^{18}\equiv 1 \mod19$

Thankfully we have $1728$ is divisible by all $6$ , $\,12$ and $18$ .

So we have $\displaystyle 2^{1728}\equiv 1 \mod7$

$\displaystyle 2^{1728}\equiv 1 \mod13$

$\displaystyle 2^{1728}\equiv 1 \mod19$

So using Chinese Remainder Theorem :-

we have :-

$\displaystyle 2^{1728}\equiv 1\mod1729$

Note that You can also use Euler's division algorithm to easily see why the three simultaneous congruence relations give the answer as 1 mod 1729

See this wiki.