Remainder on division with $22$

You can use binomial expansions and see that $$14^{16} = (22 - 8)^{16}$$ implies that the remainder is just the remainder when $(-8)^{16}( = 8^{16})$ is divided by $22.$ Proceeding similarly,

$8^{16} = 64^8 = (66 - 2)^8 \implies 2^8 = 256 \text{ divided by } 22 \implies \text{remainder = 14}$

Since $$ 14^2 \equiv -2 $$ so $$14^{16} \equiv (-2)^8 \equiv 16^2\equiv (-6)^2 \equiv 14$$

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or

$$ 14^2 = 22k -2 $$ so $$14^{16} = (22k-2)^8 = 22l+2^8=22l+22\cdot 11+14$$

A method that uses FLT.

Finding the remainder on dividing $14^{16}$ by $22$ is equivalent to twice that on dividing $7\cdot14^{15}$ by $11$.

By FLT, $14^{10}\equiv1\pmod{11}$ so $$7\cdot14^{15}\equiv7\cdot14^5\equiv98\cdot196\cdot196\equiv-1\cdot(-2)\cdot(-2)\equiv-4\equiv7\pmod{11}$$ hence the required remainder is $7\times 2=14$.