Reducing exponents mod n

Question

Say I have 5^(52) mod 22

How do I reduce this?

I know if 22 was a prime number, then I could simply I could simply reduce 5^(22) * 5^(22) * 5^(8) which would become 5^(8) mod 22 but that doesn't work in this case.

Answer 1

Hint: $5^2 \equiv 3$, and $3^3 \equiv 5$, which means $5^{6}\equiv 5$.

Answer 2

$\begin{eqnarray}{\bf Hint} &&{\rm mod}\ 11\!:\ 5^{10}\equiv 1\ \ \rm by\ \ Fermat's\ \ little\ Theorem\\ &&{\rm mod}\ 2\!:\ \ \ 5^{10}\equiv 1\ \ {\rm by}\ \ 5\equiv 1\\ \Rightarrow&&{\rm mod}\ 22\!:\ \color{#c00}{5^{10}\equiv 1}\,\Rightarrow\,\,5^{52} = 5^2 \color{#c00}{(5^{10})}^5 \equiv 25\color{#c00}{(1)}^5\equiv 3\end{eqnarray}$