How to prove $14^{11}\pmod {23}$ is same as $22 \pmod {23}$ How do we simplify the exponent in:

$$14^{11}\pmod {23} \rightarrow 22 \pmod {23} ?$$
We have that

$$14^{2}\equiv 196 \equiv 12 \mod 23$$

$$14^{4}\equiv 144 \equiv 6 \mod 23$$

and therefore

$$14^{10}\equiv 12\cdot 6^2 \equiv 432 \equiv -5 \mod 23$$

hence

$$14^{11}\equiv -70 \equiv -1 \mod 23$$
Use repeated squaring. I'll abbreviate $23|a-b$ as $a=b$. Since $14^2=12$, $14^4=12^2=6$ and $14^8=6^2=13$. But $14^3=12\times14=7$, so $14^{11}=13\times7=22$.