How to find remainder when $ 975^{40153}$ is divided by $14$? I still find tricky this kind of problems.
I tried to do solve it by factoring $14$ in $2*7$.

Then, with Fermat's Little Theorem, I find that:

$975^6\equiv 1\pmod 7$

$975^1\equiv 1\pmod 2$

How can I proceed frome here?
$\color\red{975^{3}}\equiv\color\red{1}\pmod{14}\implies$

$975^{40153}\equiv975^{3\cdot13384+1}\equiv(\color\red{975^{3}})^{13384}\cdot975^{1}\equiv\color\red{1}^{13384}\cdot975^{1}\equiv975\equiv9\pmod{14}$
$$975\equiv-5\pmod{14}\implies975^{40153}\equiv(-5)^{40153}$$

As $(-5)^3\equiv1\pmod{14}$ 

and as $4+0+1+5+3\equiv1\pmod3\implies40153\equiv1\pmod3$

$(-5)^{40153}\equiv(-5)^1\pmod{14}\equiv-5+14=?$