How to find remainder modulo $n$, when $n$ is a large number I am doing RSA questions and I really could use help! Can someone show me a simple way to find $25^9 \pmod{33}$?

In the case of $25^9 \pmod{33}$ it is pretty straightforward to use exponentiation by squaring because $9 = 8 + 1 = 2^3 + 2^0$.
$25 \equiv 25 \pmod{33}$
$25^2 \equiv 31 \pmod{33}$
$25^4 \equiv 31^2 \equiv 4 \pmod{33}$
$25^8 \equiv 4^2 \equiv 16 \pmod{33}$
$25^9 \equiv 25^8 \times 25 \equiv 16 \times 25 \equiv 4 \pmod{33}$
