jmc

We begin by replacing $b$ with $a-1$, and expressing both sides in base 10: $$\begin{array}{rl}(a-1{)}^2\cdot a^0& =7\cdot a^1+1\cdot a^0\\ \Rightarrow a^2-2a+1& =7a+1\\ \Rightarrow a^2-9a& =0\\ \Rightarrow a(a-9)& =0\end{array}$$Thus, $a$ is either 0 or 9. However, since we can't have a base equal to 0, we see that $a$ must be $\boxed{9}$.