jmc

Let's find the cycle of units digits of $7^n$, starting with $n=1$ : $7,9,3,1,7,9,3,1,\dots$ . The cycle of units digits of $7^n$ is 4 digits long: 7, 9, 3, 1. Thus, to find the units digit of $7^n$ for any positive $n$, we must find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the units digit 7, $R=2$ corresponds to the units digit 9, etc.) Since $53÷4=13R1$, the units digit of $7^{53}$ is $\boxed{7}$.