Adding 3, and still no change? Strange!

A number, when divided by 11, gives a remainder $K$. If the digit $3$ is placed at the end of the number, the remainder when divided by 11 still remains $K$.

Find the remainder when $\displaystyle K^{(K-1)^{(K-2)^{\dots^{3^{2^1}}}}}$ is divided by $11$.

Details and assumptions:-

The number is in base 10.

Placing the digit 5 at the end of $123$ makes the number into $1235$.

If $K$ was $5$, you had to find the remainder when $5^{4^{3^{2^1}}}$ is divided by $11$.

This is a part of the set 11≡ awesome (mod remainders)