Solution
= Solution
<Fermat's little theorem> reduces the exponent modulo $22$ because $23$ is a <prime number> and $5$ is <coprime> to it. Compute $5^2\equiv3\pmod{22}$, then $5^4\equiv9\pmod{22}$ and $5^5\equiv45\equiv1\pmod{22}$. Thus $5^5=22k+1$ for an <integer> $k$, and
$$
\boxed{5^{5^5}=5^{22k+1}\equiv(5^{22})^k5\equiv5\pmod{23}.}
$$
The exponent is $5^5=3125$, rather than an iterated left-associated power.