jmc

Let's find the cycle of units digits of $3^n$, starting with $n=1$ (note that the tens digit 2 in 23 has no effect on the units digit): $3,9,7,1,3,9,7,1,\dots$ . The cycle of units digits of $23^n$ is 4 digits long: 3, 9, 7, 1. Thus, to find the units digit of $23^n$ for any positive $n$, we must find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the units digit 3, $R=2$ corresponds to the units digit 9, etc.) Since $23÷4=5R3$, the units digit of $23^{23}$ is $\boxed{7}$.