Note: the following argument assumes $b=1$. To generalise, add $\ln b$ to each $\ln U_i$ term, i.e. $-n\ln b$ to $y$ so its pdf shifts.
You probably already worked out $-\ln U_i\sim\operatorname{Exp}(1)$, because $$P(-\ln U\le x)=P(U\ge\exp -x)=1-\exp -x.$$Of course, this implies $-\ln U_i$ has characteristic function $1/(1-it)$, so $Y$ has cf $1/(1-it)^n$. Now, what distribution is that? Spoiler: it's
> a Gamma distribution with $k=n,\,\theta=1$, so the pdf is $\frac{y^{n-1}}{(n-1)!}\exp -y$ for $y\ge 0$.
If you can guess that by thinking about integrals, the inversion theorem guarantees it's right because it gives the desired cf.